258 results

probability - the likelihood of an event happening. This value is always between 0 and 1.

Formula: P(Event Happening) = Number of Ways the Even Can Happen / Total Number of Outcomes

1 Die Roll

Free 1 Die Roll Calculator - Calculates the probability for the following events in the roll of one fair dice (1 dice roll calculator or 1 die roll calculator):

* Probability of any total from (1-6)

* Probability of the total being less than, less than or equal to, greater than, or greater than or equal to (1-6)

* The total being even

* The total being odd

* The total being a prime number

* The total being a non-prime number

* Rolling a list of numbers i.e. (2,5,6)

* Simulate (n) Monte Carlo die simulations.

1 die calculator

* Probability of any total from (1-6)

* Probability of the total being less than, less than or equal to, greater than, or greater than or equal to (1-6)

* The total being even

* The total being odd

* The total being a prime number

* The total being a non-prime number

* Rolling a list of numbers i.e. (2,5,6)

* Simulate (n) Monte Carlo die simulations.

1 die calculator

2 coins are tossed. Find the probability of getting 1 head and 1 tail

2 coins are tossed. Find the probability of getting 1 head and 1 tail
We can either flip HT or TH. Let's review probabilities:
[LIST]
[*]HT = 1/2 * 1/2 = 1/4 <-- We multiply since each event is independent
[*]TH = 1/2 * 1/2 = 1/4 <-- We multiply since each event is independent
[/LIST]
P(1 H, 1 T) = P(HT) + P(TH)
P(1 H, 1 T) = 1/4 + 1/4
P(1 H, 1 T) = 2/4
[URL='https://www.mathcelebrity.com/fraction.php?frac1=2%2F4&frac2=3%2F8&pl=Simplify']Using our fraction simplifier[/URL], we can reduce 2/4 to 1/2
P(1 H, 1 T) = [B]1/2[/B]

2 dice are rolled what is the probability that doubles are rolled less than 11

2 dice are rolled what is the probability that doubles are rolled less than 11
List out the doubles with a sum less than 11:
(1, 1)
(2, 2)
(3, 3)
(4, 4)
(5, 5)
The probability of each double is 1/6 * 1/6 = 1/36.
We have 5 of them, so we have 5*1/36 = [B]5/36[/B]

2 dice roll

Free 2 dice roll Calculator - Calculates the probability for the following events in a pair of fair dice rolls:

* Probability of any sum from (2-12)

* Probability of the sum being less than, less than or equal to, greater than, or greater than or equal to (2-12)

* The sum being even

* The sum being odd

* The sum being a prime number

* The sum being a non-prime number

* Rolling a list of numbers i.e. (2,5,6,12)

* Simulate (n) Monte Carlo two die simulations. 2 dice calculator

* Probability of any sum from (2-12)

* Probability of the sum being less than, less than or equal to, greater than, or greater than or equal to (2-12)

* The sum being even

* The sum being odd

* The sum being a prime number

* The sum being a non-prime number

* Rolling a list of numbers i.e. (2,5,6,12)

* Simulate (n) Monte Carlo two die simulations. 2 dice calculator

2 fair sided die are rolled. How many ways can the dice be rolled to sum exactly 6?

2 fair sided die are rolled. How many ways can the dice be rolled to sum exactly 6?
[URL='https://www.mathcelebrity.com/2dice.php?gl=1&pl=6&opdice=1&rolist=+2%2C3%2C9%2C10&dby=2%2C3%2C5&ndby=4%2C5&montect=+100']Using our 2 dice calculator[/URL], we get the following options:
[LIST]
[*]2,4
[*]3,3
[*]4,2
[*]1,5
[*]5,1
[/LIST]
The probability of rolling a sum of 6 is [B]5/36[/B]

20% of Kay’s pencils in her pencil case are broken. What is the chance of taking a pencil that is no

20% of Kay’s pencils in her pencil case are broken. What is the chance of taking a pencil that is not broken from the case if she picks one at random?
This means 100% - 20% = 80% of pencils are not broken
So the probability of drawing a pencil which is [I]not broken[/I] is 80%

2755 students were surveyed: 896 chose a falcon, 937 chose a ram, and 842 chose a panther. The remai

2755 students were surveyed: 896 chose a falcon, 937 chose a ram, and 842 chose a panther. The remaining did not vote. What is the probability that the students choice was a panther?
P(Panther) = Panther Choices / Total Students
P(Panther) = 842/2,755
P(Panther) = [B]0.3056[/B]

52 card deck what is the probability of being dealt a two

52 card deck what is the probability of being dealt a two?
There are four 2's in a deck, so we have 4/52.
4/52 is reducible to [B]1/13[/B]

6 red marbles 9 green marbles and 5 blue marbles two marbles are drawn without replacement what is t

6 red marbles 9 green marbles and 5 blue marbles two marbles are drawn without replacement what is the probability of choosing a green and then a blue marble
First draw:
there are 6 red + 9 green + 5 blue = 20 marbles
We draw 9 possible green out of 20 total marbles = 9/20
Second draw:
We don't replace, so we have 6 red + 8 green + 5 blue = 19 marbles
We draw 5 possible blue of out 19 total marbles = 5/19
Our total probability, since each event is independent, is:
[URL='https://www.mathcelebrity.com/fraction.php?frac1=9%2F20&frac2=5%2F19&pl=Multiply']9/20 * 5/19[/URL] = [B]9/76[/B]

6 sided die probability to roll a odd number or a number less than 6

6 sided die probability to roll a odd number or a number less than 6
First, we'll find the set of rolling an odd number. [URL='https://www.mathcelebrity.com/1dice.php?gl=1&opdice=1&pl=Odds&rolist=+2%2C3%2C4&dby=+2%2C3%2C5&montect=+100']From this dice calculator[/URL], we get:
Odd = {1, 3, 5}
Next, we'll find the set of rolling less than a 6. [URL='https://www.mathcelebrity.com/1dice.php?gl=4&pl=6&opdice=1&rolist=+2%2C3%2C4&dby=+2%2C3%2C5&montect=+100']From this dice calculator[/URL], we get:
Less than a 6 = {1, 2, 3, 4, 5}
The question asks for [B]or[/B]. Which means a Union:
{1, 3, 5} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5}
This probability is [B]5/6[/B]

7 black shirts 5 white shirts 10 gray shirts one is chosen at random, what is the probability that i

7 black shirts 5 white shirts 10 gray shirts one is chosen at random, what is the probability that it is not gray
[U]Find the total shirts:[/U]
Total shirts = Black Shirts + White Shirts + Gray Shirts
Total shirts = 7 + 5 + 10
Total shirts = 22
[U]Calculate the probability of choosing a gray shirt:[/U]
P(Gray) = Number of Gray shirts / Total Shirts
P(Gray) = 10/22
We can simplify this fraction. We [URL='https://www.mathcelebrity.com/fraction.php?frac1=10%2F22&frac2=3%2F8&pl=Simplify']type in 10/22 into our search engine, choose simplify[/URL], and we get:
P(Gray) = [B]5/11[/B]

a 12 sided die is rolled find the probability of rolling a number greater than 7

a 12 sided die is rolled find the probability of rolling a number greater than 7
We assume this is a fair die, not loaded.
This means each side 1-12 has an equal probability of 1/12 of being rolled.
The problem asks, P(Roll > 7)
Greater than 7 means our sample space is {8, 9, 10, 11, 12}
If each of these 5 faces have an equal probability of being rolled, then we have:
P(Roll > 7) = P(Roll = 8) + P(Roll = 9) + P(Roll = 10) + P(Roll = 11) + P(Roll = 12)
P(Roll > 7) = 1/12 + 1/12 + 1/12 + 1/12 + 1/12
P(Roll > 7) =[B] 5/12[/B]

A 12-sided die is rolled. The set of equally likely outcomes is {1,2,3,4,5,6,7,,8,9,10,11,12}. Find

A 12-sided die is rolled. The set of equally likely outcomes is {1,2,3,4,5,6,7,,8,9,10,11,12}. Find the probability of rolling a number less than 6.
We want a {1, 2, 3, 4, 5}
P(X < 6) =[B] 5/12[/B]

A 12-sided die is rolled. The set of equally likely outcomes is {1,2,3,4,5,6,7,8,9,10,11,12}. Find t

A 12-sided die is rolled. The set of equally likely outcomes is {1,2,3,4,5,6,7,8,9,10,11,12}. Find the probability of rolling a number less than 6.
We have 12 outcomes.
Less than 6 means 1, 2, 3, 4, 5.
Our probability P(x < 6) is:
P(x < 6) = [B]5/12[/B]

A 6-sided die is rolled once. What is the probability of rolling a number less than 4?

A 6-sided die is rolled once. What is the probability of rolling a number less than 4?
Using our [URL='https://www.mathcelebrity.com/1dice.php?gl=4&pl=4&opdice=1&rolist=+2%2C3%2C4&dby=+2%2C3%2C5&montect=+100']one dice calculator[/URL], we get:
P(N < 4) = [B]1/2[/B]

A bag contains 10 red balls, 10 green balls and 6 white balls. Two balls are drawn at random from th

A bag contains 10 red balls, 10 green balls and 6 white balls. Two balls are drawn at random from the bag without replacement. What is the probability that they are of different colours?
[LIST]
[*]The key phrase here is [I]without replacement[/I].
[*]First, it's easier to find the probability of both colors matching, and then subtracting that from 1.
[/LIST]
We want 1 - (P(Red-Red) + P(Green-Green) + P(White-White)). So we have the following:
[U]Find the probability of both colors matching[/U]
P(Red-Red) = 10/26 * 9/25 = 0.138462
P(Green-Green) = 10/26 * 9/25 = 0.138462
P(White-White) = 6/26 * 5/25 = 0.046154
P(Red-Red) + P(Green-Green) + P(White-White) = 0.13846 + 0.13846 + 0.046154 = 0.3231
Now, we want to take the complement of this probability which is no colors matching, so we have:
P(Both Different Colors) = 1 - P(Same Colors)
P(Both Different Colors) = 1 - 0.3231
P(Both Different Colors) = [B]0.6769[/B]

A bag contains 19 balls numbered 1 through 19. What is the probability that a randomly selected ball

A bag contains 19 balls numbered 1 through 19. What is the probability that a randomly selected ball has an even number?
Even numbers in the bag are {2,4,6,8,10,12,14,16,18}
So we have 9 total even numbers.
Therefore, the probability of drawing an even number is [B]9/19[/B]

A bag contains 2 red marbles, 3 blue marbles, and 4 green marbles. What is the probability of choosi

A bag contains 2 red marbles, 3 blue marbles, and 4 green marbles. What is the probability of choosing a blue marble, replacing it, drawing a green marble, replacing it, and then drawing a red marble?
Calculate total marbles in the bag:
Total marbles in the bag = Red Marbles + Blue Marbles + Green Marbles
Total marbles in the bag = 2 + 3 + 4
Total marbles in the bag = 9
[U]First choice, blue marble[/U]
P(blue) = Total Blue Marbles / Total Marbles in the bag
P(blue) = 3/9
[URL='https://www.mathcelebrity.com/fraction.php?frac1=3%2F9&frac2=3%2F8&pl=Simplify']Using our fraction simplifier[/URL], we see:
P(blue) = 1/3
[U]Second choice, green marble with all the marbles back in the bag after replacement[/U]
P(green) = Total Green Marbles / Total Marbles in the bag
P(green) = 4/9
[U]Third choice, red marble with all the marbles back in the bag after replacement[/U]
P(red) = Total Red Marbles / Total Marbles in the bag
P(red) = 2/9
Since each event is independent, we multiply each probability:
P(blue, green, red) = P(blue) * P(green) * P(red)
P(blue, green, red) = 1/3 * 4/9 * 2/9
P(blue, green, red) = [B]8/243[/B]

A bag contains 3 black, 4 red, 3 yellow, and 2 green marbles. What is the probability of drawing a b

A bag contains 3 black, 4 red, 3 yellow, and 2 green marbles. What is the probability of drawing a black and then a red marble out of the bag without replacing the black marble before drawing the red marble?
The phrase [U][B]without replacement[/B][/U] is a huge clue on this problem.
Take each draw and calculate the probability.
Draw 1: P(Drawing a red)
P(Drawing a red) = Total Red marbles n the jar / Total marbles in the jar
P(Drawing a red) = 4/12
4/12 simplifies to 1/3 using a common factor of 4:
P(Drawing a red) = 1/3
Draw 2: P(Drawing a black)
P(Drawing a black) = Total Black marbles in the jar / Total marbles in the jar
[I]We drew one red marble already. Without replacement means we do not put it back. Therefore, we have 12 - 1 = 11 marbles left in the jar.[/I]
P(Drawing a black) = 3/11
The question asks, what is the the following probability:
P(Drawing a Red, Drawing a Black)
Because each draw is [I][U]independent[/U], [/I]we multiply each draw probability together:
P(Drawing a Red, Black) = P(Drawing a Red) * P(Drawing a Black)
P(Drawing a Red, Black) = 1/3 * 3/11
P(Drawing a Red, Black) = [B]1/11[/B]

A bag contains 3 red marbles and 4 blue marbles. a marble is taken at random and replaced. another m

A bag contains 3 red marbles and 4 blue marbles. a marble is taken at random and replaced. Another marble is taken from the bag. Work out the probability that the two marbles taken from the bag are the same color.
[LIST]
[*]Total number of marbles in the bag is 3 + 4 = 7.
[*]The problem asks for the probability of (RR) [I]or[/I] (BB).
[*]It's worthy to note we are replacing the balls after each draw, which means we always have 7 to draw from
[/LIST]
Since each draw is independent, we take the product of each event for the total event probability.
P(RR) = 3/7 * 3/7 = 9/49
P(BB) = 4/7 * 4/7 = 16/49
We want to know P(RR) + P(BB)
P(RR) + P(BB) = 9/49 + 16/49 = 25/49
[MEDIA=youtube]26F9vjsgNGs[/MEDIA]

A bag contains 3 white balls and 2 black balls. Another bag contains 2 white and 4 black balls. A ba

A bag contains 3 white balls and 2 black balls. Another bag contains 2 white and 4 black balls. A bag and a ball are picked random. Then probability of that the ball will be white is:
Probability that you pick any bag is 0.5.
Bag 1 White Ball = 0.5(3/5) = 3/10 = 0.3
Bag 2 White Ball = 0.5(2/6) = 1/6 = 0.16667
Add them both:
0.3 + 0.16667 = [B]0.46667[/B]

A bag contains 6 red balls and 7 green balls. You plan to select 4 balls at random. Determine the pr

A bag contains 6 red balls and 7 green balls. You plan to select 4 balls at random. Determine the probability of selecting 4 green balls.
Assuming draw without replacement of the balls, we have:
[LIST=1]
[*]Selection 1: 7 green out of 13 balls
[*]Selection 2: 6 green out of 12 balls
[*]Selection 3: 5 green out of 11 balls
[*]Selection 4: 4 green out of 10 balls
[/LIST]
Since each draw is independent, we multiply each probability of green:
P(GGGG) = 7/13 * 6/12 * 5/11 * 4/10
P(GGGG) = 840/17,160
P(GGGG) = [B]0.05[/B]

A bag contains 666 red balls, 444 green balls, and 333 blue balls. If we choose a ball, then another

A bag contains 666 red balls, 444 green balls, and 333 blue balls. If we choose a ball, then another ball without putting the first one back in the bag, what is the probability that the first ball will be green and the second will be red?
[U]Calculate total number of balls to start:[/U]
Total Balls = Red Balls + Green Balls + Blue Balls
Total Balls = 666 + 444 + 333
Total Balls = 1,443
[U]Calculate the probability of drawing a green ball on the first pick:[/U]
P(Green) = Green Balls / Total Balls
P(Green) = 444/1443
P(Green) = 0.30769
[U]Calculate the probability of drawing a red ball on the second pick (without replacement):[/U]
Total Balls decrease by 1, since we do not replace. So Total Balls = 1,443 - 1 = 1,442
P(Red) = Red Balls / Total Balls
P(Red) = 666/1442
P(Red) = 0.46186
Now, we want the probability of Green, Red in that order.
Since each event is independent, we multiply the event probabilities
P(Green, Red) = P(Green) * P(Red)
P(Green, Red) = 0.30769 * 0.46186
P(Green, Red) = [B]0.14211[/B]

A bag contains 7 red, 9 white, and 4 blue marbles. Find the probability of picking 3 blue marbles if

A bag contains 7 red, 9 white, and 4 blue marbles. Find the probability of picking 3 blue marbles if each marble is NOT returned to the bag before the next marble is picked.
The problem states we will have no replacement.
[LIST]
[*]First draw probability is 4 blue marbles out of (7 red + 9 white + 4 blue) = 20 marbles (4/20)
[*]Second draw probability is 3 blue marbles out of (7 red + 9 white + 3 blue) = 19 marbles (3/19)
[*]Third draw probability is 2 blue marbles out of (7 red + 9 white + 2 blue) = 18 marbles (2/18)
[/LIST]
Each draw is independent, so we multiply the three draws together:
4/20 * 3/19 * 2/18
24/6840
[B]0.0035[/B]

A bag contains tiles, 3 tiles are red. 6 tiles are green, and 3 tiles are blue. A tile will be rando

A bag contains tiles, 3 tiles are red. 6 tiles are green, and 3 tiles are blue. A tile will be randomly selected from the bag . What is the probability that the tile selected will be green
P(green) = Number of green tiles / Total Tiles
P(green) = 6 / (3 + 6 + 3)
P(green) = 6 / 12
We can simplify this fraction. We [URL='https://www.mathcelebrity.com/fraction.php?frac1=6%2F12&frac2=3%2F8&pl=Simplify']type in 6/12 into our search engine, pick simplify[/URL], and we get:
P(green) = [B]1/2 or 0.5[/B]

A bag has 3 red, 5 blue, and 2 yellow pieces of candy. What is the theoretical probability of drawin

A bag has 3 red, 5 blue, and 2 yellow pieces of candy. What is the theoretical probability of drawing a blue piece of candy
P(Blue) = Blue Candies / Total Candies
P(Blue) = 5 / (3 + 5 + 2)
P(Blue) = 5 / 10
We can [URL='https://www.mathcelebrity.com/search.php?q=5%2F10&x=0&y=0']simplify this using our GCF calculator[/URL] and we get:
P(Blue) = [B]1/2[/B]

A binomial probability experient is conducted with the given parameters. Compute the probability of

A binomial probability experient is conducted with the given parameters. Compute the probability of x successes in the n independent trials of the experiment.
n = 40, p = 0.05, x = 2
P(2) =
Answer is [B]0.2777[/B]. Using Excel formula of =BINOMDIST(2,40,0.05,FALSE) or using our [URL='http://www.mathcelebrity.combinomial.php?n=+40&p=0.05&k=2&t=+5&pl=P%28X+%3D+k%29']binomial probability calculator[/URL]

A blue dice and a red dice are tossed what is the probability that a 6 will appear on both dice

A blue dice and a red dice are tossed what is the probability that a 6 will appear on both dice
Each event is independent.
P(Blue dice 6) = 1/6
P(Red Dice 6) = 1/6
P(Blue 6, Red 6) = 1/6 * 1/6 = [B]1/36[/B]

A blue die and a green die are rolled. Find the probability that the blue and green are both less th

A blue die and a green die are rolled. Find the probability that the blue and green are both less than 6
P(Blue not 6) = 5/6
P(Green not 6) = 5/6
Each one is independent of the other, so the probability that both are less than 6 is:
P(Both not 6) = P(Blue not 6) x P(Green not 6)
P(Both not 6) = 5/6 * 5/6
P(Both not 6) = [B]25/36 = 0.6944[/B]

A bowler knocks down at least 6 pins 70 percent of the time. Out of 200 rolls, how many times can yo

A bowler knocks down at least 6 pins 70 percent of the time. Out of 200 rolls, how many times can you predict the bowler will knock down at least 6 pins?
Expected Value of (knocking down at least 6 pins) = number of rolls * probability of knocking down at least 6 pins
Expected Value of (knocking down at least 6 pins) = 200 * 0.7
Expected Value of (knocking down at least 6 pins) = [B]140[/B]

A box

A box contains 4 plain pencils and 4 pens. A second box contains 5 color pencils and 3 crayons. One item from each box is chosen at random. What is the probability that a pen from the first box and a crayon from the second box are selected?
[LIST]
[*]First box, P(pen) = 4/8 = 1/2 = 0.5
[*]Second box, P(crayon) = 3/8
[/LIST]
Since each event is independent, we have:
P(Pen from Box 1) * P(Crayon from Box 2) = 1/2 * 3/8 = [B]3/16 or 0.1875[/B]

A box contains 10 bells. There are 6 red bells and the rest are silver. What is the probability of p

A box contains 10 bells. There are 6 red bells and the rest are silver. What is the probability of picking two bells of the same color if the bell is replaced after each pick?
If there are 6 red bells, then we have 10 - 6 = 4 silver bells.
The problem asks for the probability of picking two bells of the same color. Which mean we have 2 scenarios:
[LIST=1]
[*]Silver, Silver
[*]Red, Red
[/LIST]
Find the probability of Silver, Silver:
Since each draw is independent, and we replace the bells, we have a 4/10 chance of picking silver. [URL='https://www.mathcelebrity.com/fraction.php?frac1=4%2F10&frac2=3%2F8&pl=Simplify']Simplified, this is 2/5[/URL].
(2/5)(2/5) = 4/25
Find the probability of Red, Red:
Since each draw is independent, and we replace the bells, we have a 6/10 chance of picking silver. [URL='https://www.mathcelebrity.com/fraction.php?frac1=6%2F10&frac2=3%2F8&pl=Simplify']Simplified, this is 3/5[/URL].
(3/5)(3/5) = 9/25
Because we want Silver, Silver [B][U]or[/U][/B] Red, Red, we add the two probabilities.
4/25 + 9/25 = [B]13/25[/B]

A box contains 22 red apples and 3 green apples. Three apples are selected at random, one after the

A box contains 22 red apples and 3 green apples. Three apples are selected at random, one after the other, without replacement. please show the steps.
(a) The first two apples are green. What is the probability that the third apple is red?
(b) What is the probability that exactly two of the three apples are red?
a) You have 22 red apples left and 1 green left leaving 23 total apples left. Therefore, probability of red is
[B]P(R) = 22/23[/B]
b) Determine our sample space to select exactly two red apples in three picks.
[LIST=1]
[*]RRG
[*]RGR
[*]GRR
[/LIST]
[U]Now determine the probabilities of each event in the sample space[/U]
P(RRG) = 22/25 * 21/24 * 3/23 = 0.1004
P(RGR) = 22/25 * 3/24 * 21/23 = 0.1004
P(GRR) = 3/25 * 22/24 * 21/23 = 0.1004
[U]We want the sum of the three probabilities[/U]
P(RRG) + P(RGR) + P(GRR) = 0.1004 + 0.1004 + 0.1004
P(RRG) + P(RGR) + P(GRR) = 3(0.1004)
P(RRG) + P(RGR) + P(GRR) = [B]0.3012[/B]

A box contains 4 plain pencils and 4 pens. A second box contains 5 color pencils and 3 crayons. One

A box contains 4 plain pencils and 4 pens. A second box contains 5 color pencils and 3 crayons. One item from each box is chosen at random. What is the probability that a pen from the first box and a crayon from the second box are selected?
[LIST]
[*]First box, P(pen) = 4/8 = 1/2 = 0.5
[*]Second box, P(crayon) = 3/8
[/LIST]
Since each event is independent, we have:
P(Pen from Box 1) * P(Crayon from Box 2) = 1/2 * 3/8 = [B]3/16 or 0.1875[/B]

A box contains 4 red jellies, 6 blue jellies and 5 yellow jellies. What is the probability that a j

A box contains 4 red jellies, 6 blue jellies and 5 yellow jellies. What is the probability that a jelly chosen randomly from the box is not red?
Calculate Total Jellies:
Total Jellies = Red Jellies + Blue Jellies + Yellow Jellies
Total Jellies = 4 + 6 + 5
Total Jellies = 15
Not choosing a red jelly means choosing a blue [B]or[/B] yellow jelly
P(not red jelly) = P(blue) + P(Yellow)
P(not red jelly) = (Blue Jelly + Yellow Jelly) / Total Jellies
P(not red jelly) = (6 + 5)/15
P(not red jelly) = [B]11/15[/B]

A box contains 5 black and 2 white balls. 2 balls are drawn without replacement. Find the probabilit

A box contains 5 black and 2 white balls. 2 balls are drawn without replacement. Find the probability of drawing 2 black balls.
First draw probability of black is:
Total Balls in box = Black balls + white balls
Total Balls in Box = 5 + 2
Total Balls in Box = 7
P(Black) = Black Balls / Total balls in box
P(Black) = 5/7
Second draw probability of black (with no replacement) is:
Total Balls in box = Black balls + white balls
Total Balls in Box = 4 + 2
Total Balls in Box = 6
P(Black) = Black Balls / Total balls in box
P(Black) = 4/6
Using our [URL='https://www.mathcelebrity.com/fraction.php?frac1=4%2F6&frac2=3%2F8&pl=Simplify']fraction simplifier[/URL], we see that 4/6 is:
2/3
Since each event is independent, we can multiply them to find the probability of drawing 2 black balls:
P(Black, Black) = 5/7 * 2/3
[URL='https://www.mathcelebrity.com/fraction.php?frac1=5%2F7&frac2=2%2F3&pl=Multiply']P(Black, Black)[/URL] = 10/21
[MEDIA=youtube]HEa_G3nwgUQ[/MEDIA]

A box contains 5 plain pencils and 3 pens. A second box contains 2 color pencils and 2 crayons . One

A box contains 5 plain pencils and 3 pens. A second box contains 2 color pencils and 2 crayons. One item from each box is chosen at random. What is the probability that a plain pencil from the first box and a color pencil from the second box are selected
[U]Calculate the probability of a plain pencil in the first box:[/U]
P(plain pencil in the first box) = Total Pencils / Total Objects
P(plain pencil in the first box) = 5 pencils / (5 pencils + 3 pens)
P(plain pencil in the first box) = 5/8
[U]Calculate the probability of a color pencil in the first box:[/U]
P(color in the second box) = Total Pencils / Total Objects
P(color in the second box) = 2 pencils / (2 pencils + 2 crayons)
P(color in the second box) = 2/4
We can simplify this. [URL='https://www.mathcelebrity.com/fraction.php?frac1=2%2F4&frac2=3%2F8&pl=Simplify']Type 2/4 into our search engine[/URL] and we get 1/2
Now the problem asks for the probability that a plain pencil from the first box and a color pencil from the second box are selected.
Since each event is independent, we multiply them together to get our answer:
P(plain pencil in the first box, color in the second box) = P(plain pencil in the first box) * P(color in the second box)
P(plain pencil in the first box, color in the second box) = 5/8 * 1/2
P(plain pencil in the first box, color in the second box) = [B]5/16[/B]

A box contains 5 plain pencils and 7 pens. A second box contains 4 color pencils and 4 crayons. One

A box contains 5 plain pencils and 7 pens. A second box contains 4 color pencils and 4 crayons. One item from each box is chosen at random. What is the probability that a plain pencil from the first box and a color pencil from the second box are selected?
Probability of plain pencil from first box:
5/(5 + 7) = 5/12
Probability of color pencil from second box:
4/(4 + 4) = 4/8 = 1/2
Probability of both events together:
Since each event is independent, we multiply probabilities:
5/12 * 1/2 = [B]5/24[/B]

A box contains 6 yellow, 3 red, 5 green, and 7 blue colored pencils. A pencil is chosen at random, i

A box contains 6 yellow, 3 red, 5 green, and 7 blue colored pencils. A pencil is chosen at random, it is not replaced, then another is chosen. What is the probability of choosing a red followed by a green?
We have 6 + 3 + 5 + 7 = 21 total pencils
P(Red on the first draw) = Total Red / Total pencils
P(Red on the first draw) = 3/21
[URL='https://www.mathcelebrity.com/fraction.php?frac1=3%2F21&frac2=3%2F8&pl=Simplify']P(Red on the first draw)[/URL] = 1/7
We're drawing without replacement, this means on the next draw, we have 21 - 1 = 20 pencils
P(Green on the second draw) = Total Green / Total pencils
P(Green on the second draw) = 5/20
[URL='https://www.mathcelebrity.com/fraction.php?frac1=5%2F20&frac2=3%2F8&pl=Simplify']P(Green on the second draw) [/URL]= 1/4
Since each event is independent, we have:
P(Red on first, green on second) = P(Red on First) * P(green on second)
P(Red on first, green on second) = 1/7 * 1/4
P(Red on first, green on second) = [B]1/28[/B]

A box is filled with 10 green cards, 4 blue cards, and 4 brown cards. A card is chosen at random fr

A box is filled with 10 green cards, 4 blue cards, and 4 brown cards. A card is chosen at random from the box. What is the probability that it is a green or a brown card?
Calculate Total Cards:
10 green cards + 4 blue cards + 4 brown cards = 18 cards
"Or", means either or, so we want P(Green) + P(Brown)
[U]Find P(Green)[/U]
P(Green) = Green Cards / Total Cards
P(Green) = 10/18 <-- Simplify by dividing top and bottom by 2
P(Green)= 5/9 <-- Simplify by dividing top and bottom by 2
Find P(Brown)
P(Brown) = Brown Cards / Total Cards
P(Brown) = 4/18 <-- Simplify by dividing top and bottom by 2
P(Brown)= 2/9 <-- Simplify by dividing top and bottom by 2
P(Green) + P(Brown) = 5/9 + 2/9
P(Green) + P(Brown) = [B]7/9[/B]

A box is filled with 5 blue cards,2 red cards, and 5 yellow cards. A card is chosen at random from t

A box is filled with 5 blue cards,2 red cards, and 5 yellow cards. A card is chosen at random from the box. What is the probability that it is a blue or a yellow card? Write your answer as a fraction in simplest form.
We want P(B) + P(Y)
P(B) = 5/12
P(Y) = 5/12
P(B) + P(Y) = 5/12 + 5/12 = 10/12
Reduce this fraction using 2 as our common factor:
[B]5/6[/B]

A candidate for mayor wants to gauge potential voter reaction to an increase recreational services b

A candidate for mayor wants to gauge potential voter reaction to an increase recreational services by estimating the proportion of voter who now use city services. If we assume that 50% of the voters require city recreational services, what is the probability that 40% or fewer voters in a sample of 100 actually will use these city services?
First, let's do a test on the proportion using our [URL='http://www.mathcelebrity.com/proportion_hypothesis.php?x=+40&n=+100&ptype=%3D&p=+0.5&alpha=+0.05&pl=Proportion+Hypothesis+Testing']proportion hypothesis calculator[/URL]:
We get Z = -2
Now use the [URL='http://www.mathcelebrity.com/zscore.php?z=p%28z%3C-2%29&pl=Calculate+Probability']z-score calculator[/URL] to get P(z<-2) = [B]0.02275[/B]

a card is chosen at a random from a deck of 52 cards. it is then replaced and a second card is chose

a card is chosen at a random from a deck of 52 cards. it is then replaced and a second card is chosen. what is the probability of getting a jack and then an eight?
Calculate the probability of drawing a jack from a full deck
There are 4 jacks in a deck of 52 cards
P(J) = 4/52
P(J) = 1/13 <-- We simplify 4/52 by dividing top and bottom of the fraction by 4
Calculate the probability of drawing an eight from a full deck
There are 4 eights in a deck of 52 cards. We[I] replaced[/I] the first card giving us 52 cards to choose from.
P(8) = 4/52
P(8) = 1/13 <-- We simplify 4/52 by dividing top and bottom of the fraction by 4
Since each event is independent, we multiply:
P(J, 8) = P(J) * P(8)
P(J, 8) = 1/13 * 1/13
P(J, 8) = [B]1/169[/B]

a card is drawn at random from a standard 52 card deck. find the probability that the card is not a

a card is drawn at random from a standard 52 card deck. find the probability that the card is not a king.
There are 4 kings in a standard 52 card deck. To not get a king, we'd have 52 - 4 = 48 possible cards.
The probability of not drawing a King is 48/52.
But we can simplify this. So we [URL='https://www.mathcelebrity.com/fraction.php?frac1=48%2F52&frac2=3%2F8&pl=Simplify']type the fraction 48/52 into our search engine[/URL], and get:
[B]12/13[/B]

A card is drawn from a pack of 52 cards. The probability that the card drawn is a red card is

A card is drawn from a pack of 52 cards. The probability that the card drawn is a red card is
The deck is split evenly between red and black cards.
So we have 52/2 = 26 red cards
P(Red) = # of Red Cards / Total Deck Cards
P(Red) = 26/52
We can simplify this fraction. [URL='https://www.mathcelebrity.com/fraction.php?frac1=26%2F52&frac2=3%2F8&pl=Simplify']Using our fraction calculator[/URL], we get:
P(Red) = [B]1/2 or 0.5[/B]

A card is drawn from a standard deck of 52 cards. What is the probability of drawing an ace or a 6

A card is drawn from a standard deck of 52 cards. What is the probability of drawing an ace or a 6?
There are 4 Ace's in a standard 52 card deck. There are 4 6's as well.
So we have 4 + 4 = 8 possible cards out of 52:
8/52
To simplify, [URL='https://www.mathcelebrity.com/fraction.php?frac1=8%2F52&frac2=3%2F8&pl=Simplify']we type this into our search engine[/URL] and we get:
[B]2/13[/B]

A card is picked from a deck of 52 cards. Find the probability of getting a black ace or a red queen

A card is picked from a deck of 52 cards. Find the probability of getting a black ace or a red queen.
In a standard deck of 52 cards, we have:
[LIST]
[*]2 black Aces with probability 2/52 = 1/26 [URL='https://www.mathcelebrity.com/fraction.php?frac1=2%2F52&frac2=3%2F8&pl=Simplify']using our fraction simplifier[/URL]
[*]2 red Queens with probability 2/52 = 1/26 [URL='http://using our fraction simplifier']using our fraction simplifier[/URL]
[/LIST]
The problems asks for P(Red Queen Or Black Ace). Or means we add, so we have:
P(Red Queen Or Black Ace) = P(Red Queen) + P(Black Ace)
P(Red Queen Or P Black Ace) = 1/26 + 1/26
P(Red Queen Or P Black Ace) = 2/26
P(Red Queen Or P Black Ace) = [B]1/13[/B] [URL='https://www.mathcelebrity.com/fraction.php?frac1=2%2F26&frac2=3%2F8&pl=Simplify']using our fraction simplifier[/URL]

a card selected from a deck of 52 cards what is the probability it is a black card or face card

a card selected from a deck of 52 cards what is the probability it is a black card or face card
Facts:
[LIST]
[*]Half the cards in the deck are black (26/52)
[*]There are 12 face cards (K, Q, J) in a deck (12/52)
[*]Black and Face = 6/52 (Duplicates from above)
[/LIST]
P(Black or Face) = P(Black) + P(Face) - P(Black And Face)
P(Black or Face) = 26/52 + 12/52 - 6/52
P(Black or Face) = 32/52
We can simplify this. We use our [URL='https://www.mathcelebrity.com/fraction.php?frac1=32%2F52&frac2=3%2F8&pl=Simplify']fraction simplifier[/URL] to get:
P(Black or Face) = [B]8/13[/B]

a cash prize of $4600 is to be awarded at a fundraiser. if 2300 tickets are sold at $7 each, find th

a cash prize of $4600 is to be awarded at a fundraiser. if 2300 tickets are sold at $7 each, find the expected value.
Expected Value E(x) is:
E(x) = Probability of winning * Winning Price - Probability of losing * Ticket Price
[U]Since only 1 cash price will be given, 2299 will be losers:[/U]
E(x) = 4600 * (1/2300) - 2299/2300 * 7
E(x) = 2 - 0.99956521739 * 7
E(x) - 2 - 7
E(x) = [B]-5[/B]

A certain group of woman has a 0.69% rate of red/green color blindness. If a woman is randomly selec

A certain group of woman has a 0.69% rate of red/green color blindness. If a woman is randomly selected, what is the probability that she does not have red/green color blindness?
0.69% = 0.0069.
There exists a statistics theorem for an event A that states:
P(A) + P(A') = 1 where A' is the event not happening
In this case, A is the woman having red/green color blindness. So A' is the woman [U][B][I]not[/I][/B][/U][I] having red/green color blindness[/I]
So we have:
0.0069 + P(A') = 1
Subtract 0.0069 from each side, we get:
P(A') = 1 - 0.0069
P(A') = [B]0.9931[/B]

A class is made up of 6 boys and 12 girls. Half of the girls wear glasses. A student is selected at

A class is made up of 6 boys and 12 girls. Half of the girls wear glasses. A student is selected at random from the class. What is the probability that the student is a girl with glasses?
1/2 of 6 is 3. So we want the probability we pick any of the 3 girls wearing glasses.
We have a total of 6 + 12 = 18 people.
Our probability is [B]3/18, or 1/6[/B].

A coffee franchise is opening a new store. The company estimates that there is a 75% chance the sto

A coffee franchise is opening a new store. The company estimates that there is a 75% chance the store will have a profit of $45,000, a 10% chance the store will break even, and a 15% chance the store will lose $2,500. Determine the expected gain or loss for this store.
Calculate the expected value E(x). Expected value is the sum of each event probability times the payoff or loss:
E(x) = 0.75(45,000) + 0.1(0) + 0.15(-2,500) <-- Note, break even means no profit and no loss and a loss is denoted with a negative sign
E(x) = 33,750 + 0 - 375
E(x) = [B]33,375 gain[/B]

A coin is tossed 3 times. a. Draw a tree diagram and list the sample space that shows all the possib

A coin is tossed 3 times. a. Draw a tree diagram and list the sample space that shows all the possible outcomes
[URL='https://www.mathcelebrity.com/cointoss.php?hts=+HTHTHH&hct=+2&tct=+1&fct=+5>=no+more+than&nmnl=+2&htpick=heads&tossct=3&montect=3&calc=5&pl=Calculate+Probability']type in "toss a coin 3 times" and pick the probability option[/URL].

A coin is tossed and a die is rolled. Find the probability pf getting a head and a number greater th

A coin is tossed and a die is rolled. Find the probability pf getting a head and a number greater than 4.
Since each event is independent, we multiply the probabilities of each event.
P(H) = 0.5 or 1/2
P(Dice > 4) = P(5) + P(6) = 1/6 + 1/6 = 2/6 = 1/3
P(H) AND P(Dice > 4) = 1/2 * 1/3 = [B]1/6
[MEDIA=youtube]ofsbmHmQmjs[/MEDIA][/B]

A compact disc is designed to last an average of 4 years with a standard deviation of 0.8 years. Wha

A compact disc is designed to last an average of 4 years with a standard deviation of 0.8 years. What is the probability that a CD will last less than 3 years?
Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=3&mean=4&stdev=0.8&n=1&pl=P%28X+%3C+Z%29']Z-score and Normal distribution calculator[/URL], we get:
[B]0.10565[/B]

A computer randomly generates a whole number from 1 to 25. Find the probability that the computer ge

A computer randomly generates a whole number from 1 to 25. Find the probability that the computer generates a multiple of 5
[URL='https://www.mathcelebrity.com/factoriz.php?num=25&pl=Show+Factorization']Multiples of 5[/URL]:
{1, 5, 25}
So we have the probability of a random number multiple of 5 is
[B]3/25[/B]

a cookie jar contains 3 vanilla, 2 chocolate chip, and 7 gingersnap cookies. Of one cookie is taken

a cookie jar contains 3 vanilla, 2 chocolate chip, and 7 gingersnap cookies. Of one cookie is taken at random from the jar, what is the probability that it will be a vanilla cookie?
Total cookies = 3 + 2 + 7
Total cookies = 12
P(Vanilla) = Vanilla Cookies / Total Cookies
P(Vanilla) = 3/12
Simplified dividing top and bottom by 3, we have:
P(Vanilla) = [B]1/4[/B]

A desk drawer contains 10 blue pencils, 7 red pencils, and 8 green pencils. Without looking, you dra

A desk drawer contains 10 blue pencils, 7 red pencils, and 8 green pencils. Without looking, you draw out a pencil and then draw out a second pencil without returning the first pencil. What is the probability that the first pencil and the second pencil are both green?
We are drawing without replacement. Take each draw probability:
[LIST=1]
[*]First draw, we have a total of 10 + 7 + 8 = 25 pencils to choose from. P(Green) = 8/25
[*]Next draw, we only have 24 total pencils, and 7 green pencils since we do not replace. Therefore, we have P(Green)= 7/24
[/LIST]
Since both events are independent, we have:
P(Green) * P(Green) = 8/25 * 7/24
P(Green) * P(Green) = 56/600
Using our [URL='http://www.mathcelebrity.com/gcflcm.php?num1=56&num2=600&num3=&pl=GCF']GCF Calculator[/URL], we see the greatest common factor of 56 and 600 is 8. So we divide top and bottom of the fraction by 8.
[B]P(Green) * P(Green) = 7/75[/B]

A dice has six sides. The dice is rolled once. What is the probability that a six will be the result

A dice has six sides. The dice is rolled once. What is the probability that a six will be the result.
P(6) = [B]1/6[/B]

A die and a coin are tossed. What is the probability of getting a 6 and a tail?

A die and a coin are tossed. What is the probability of getting a 6 and a tail?
Roll a 6:
P(6) = 1/6
Flip a tail:
P(T) = 1/2
Probability of getting a 6 and a tail:
Since both events are independent, we have:
P(6 and T) = P(6) * P(T)
P(6 and T) = 1/6 * 1/2
P(6 and T) = [B]1/12[/B]

A drawer is filled with 9 black shirts , 6 white shirts, and 5 gray shirts one shirt is chosen at ra

A drawer is filled with 9 black shirts , 6 white shirts, and 5 gray shirts one shirt is chosen at random from the drawer find the probability that it is not a white shirt
P(Not White) = P(Black or Gray)
P(Black or Gray) = (Total Black + Total Gray)/Total Shirts
P(Black or Gray) = (9 + 5)/(9 + 6 + 5)
P(Black or Gray) = 14/20
Simplifying this [URL='https://www.mathcelebrity.com/fraction.php?frac1=14%2F20&frac2=3%2F8&pl=Simplify']using our fraction simplify calculator[/URL], we get:
P(Black or Gray) = [B]7/10, or 0.7 or 70%[/B]

A fair coin is tossed 4 times. a) How many outcomes are there in the sample space? b) What is the pr

A fair coin is tossed 4 times.
a) How many outcomes are there in the sample space?
b) What is the probability that the third toss is heads, given that the first toss is heads?
c) Let A be the event that the first toss is heads, and B be the event that the third toss is heads. Are A
and B independent? Why or why not?
a) 2^4 = [B]16[/B] on our [URL='http://www.mathcelebrity.comcointoss.php?hts=+HTHTHH&hct=+2&tct=+1&fct=+5>=no+more+than&nmnl=+2&htpick=heads&tossct=+4&calc=5&montect=+500&pl=Calculate+Probability']coin toss calculator[/URL]
b) On the link above, 4 of those outcomes have H and H in toss 1 and 3. So it's [B]1/4 or 0.25[/B]
c) [B]Yes, each toss is independent of each other.[/B]

A fair die is rolled. What is the probability of rolling a 3 or a 6?

A fair die is rolled. What is the probability of rolling a 3 or a 6?
P(3 or 6) can be written as:
P(3) + P(6)
A fair die means all faces have an equal probability of 1/6
P(3) = 1/6
P(6) = 1/6
P(3 or 6) = P(3) + P(6)
P(3 or 6) = 1/6 + 1/6
P(3 or 6) = 2/6
[URL='https://www.mathcelebrity.com/fraction.php?frac1=2%2F6&frac2=3%2F8&pl=Simplify']Using our fractions simplifier for 2/6[/URL], we get:
P(3 or 6) = [B]1/3[/B]

A fake coin has heads on both sides, if the coin tossed once, what is the probability of getting a h

A fake coin has heads on both sides, if the coin tossed once, what is the probability of getting a head?
Since you always flip a head, we have:
P(Head) = [B]1 or 100%[/B]

A financial analyst computed the ROI for all companies listed on the NYSE. She found that the mean o

A financial analyst computed the ROI for all companies listed on the NYSE. She found that the mean of this distribution was 10% with standard deviation of 5%. She is interested in examining further those companies whose ROI is between 14% and 16% of the approximately 1,500 companies listed on the exchange, how many are of interest of her?
First, use our [URL='http://www.mathcelebrity.com/zscore.php?z=p%280.14%3Cz%3C0.16%29&pl=Calculate+Probability']z-score calculator[/URL] to get P(0.14 < z < 0.16) = 0.007889
Divide that by 2 for two-tail test to get0.003944729
Use the NORMSINV(0.003944729) in Excel to get the Z value of 2.66
Therefore, the companies of interest are 2.66 * 1500 * 0.10 = [B]399[/B]

A fuel injection system is designed to last 18 years, with a standard deviation of 1.4 years. What i

A fuel injection system is designed to last 18 years, with a standard deviation of 1.4 years. What is the probability that a fuel injection system will last less than 15 years?
Using our [URL='https://www.mathcelebrity.com/probnormdist.php?xone=15&mean=18&stdev=14&n=1&pl=P%28X+%3C+Z%29']z-score calculator[/URL], we see that:
P(X < 15) = [B]0.416834[/B]

A golf ball is selected at random from a golf bag. If the golf bag contains 5 type A balls, 4 type B

A golf ball is selected at random from a golf bag. If the golf bag contains 5 type A balls, 4 type B balls, and 9 type C balls, find the probability that the golf ball is not a type A ball
Not Type A means Type B or Type C.
Our total balls = 5 + 4 + 9 = 18
P(B or C) = P(B) + P(C)
P(B or C) = 4/18 + 9/18
P(B or C) = [B]13/18[/B]

A group of 30 students from your school is part of the audience for a TV game show. The total number

A group of 30 students from your school is part of the audience for a TV game show. The total number of people in the audience is 120. What theoretical probability of 5 students from your school being selected as contestants out of 9 possible contestant spots?
We want the probability a student from your school is chosen out of total students times total ways to choose students from your school:
[U]a) P(5 students being selected):[/U]
5/30 * 4/(120 - 30)
5/30 * 4/90
20/2700
[URL='https://www.mathcelebrity.com/fraction.php?frac1=20%2F2700&frac2=3%2F8&pl=Simplify']Simplifying this fraction[/URL], we get:
1/135
[U]b) Total Ways 9 students can be picked from your school:[/U]
9/120
[URL='https://www.mathcelebrity.com/fraction.php?frac1=9%2F120&frac2=3%2F8&pl=Simplify']Simplifying this fraction[/URL], we get:
3/40
Divide a by b:
1/135 / 3/40
40/405
[URL='https://www.mathcelebrity.com/fraction.php?frac1=40%2F405&frac2=3%2F8&pl=Simplify']Simplifying[/URL], we get:
[B]8/81[/B]

A group of people were asked if they had run a red light in the last year. 497 responded "yes", and

A group of people were asked if they had run a red light in the last year. 497 responded "yes", and 223 responded "no". Find the probability that if a person is chosen at random, they have run a red light in the last year.
P(Run Red Light) = yes answers / total answers
P(Run Red Light) = 497 / (497 + 223)
P(Run Red Light) = 497 / 720
P(Run Red Light) = [B]0.6903[/B]

A gym has 18 exercise stations, including 2 rowing machines. What is the probability that a randoml

A gym has 18 exercise stations, including 2 rowing machines. What is the probability that a randomly selected exercise station will be a rowing machine?
The probability is 2/18.
We can simplify this fraction. Divide top and bottom by 2:
[B]1/9[/B]

A high school with 1000 students offers two foreign language courses : French and Japanese. There ar

A high school with 1000 students offers two foreign language courses : French and Japanese. There are 200 students in the French class roster, and 80 students in the Japanese class roster. We also know that 30 students enroll in both courses. Find the probability that a random selected student takes neither foreign language course.
Let F be the event a student takes French and J be the event a student takes Japanese
P(F) = 200/1000 = 0.2
P(J) = 80/1000 = 0.08
P(F ∩ J) = 30/1000 = 0.03
From our [URL='http://www.mathcelebrity.com/probunion2.php?pa=+0.2&pb=0.08+&paintb=+0.03&aub=+&pl=Calculate']two event calculator[/URL], we get P(F U J) = 0.25
So we want P(F U J)^C = 1 - P(F U J) = 1 - 0.25 = [B]0.75[/B]

A hypothetical population consists of eight individuals ages 14,15,17,20,26,27,28, and 30 years. Wh

A hypothetical population consists of eight individuals ages 14,15,17,20,26,27,28, and 30 years. What is the probability of selecting a participant who is at least 20 years old?
At least 20 means 20 or older, so our selection of individuals is:
{20, 26, 27, 28, 30}
This is 5 out of a possible 8, so we have [URL='http://www.mathcelebrity.com/perc.php?num=5&den=8&pcheck=1&num1=16&pct1=80&pct2=70&den1=80&idpct1=10&hltype=1&idpct2=90&pct=82&decimal=+65.236&astart=12&aend=20&wp1=20&wp2=30&pl=Calculate']5/8 of 0.625, which is 62.5%[/URL]

A jar contains 7 red marbles, 8 green marbles, and 6 blue marbles. What is the probability that you

A jar contains 7 red marbles, 8 green marbles, and 6 blue marbles. What is the probability that you draw 4 green marbles in a row if you do not replace the marbles after each draw?
The key phrase in this problem is [I]do not replace[/I].
[U]Draw #1:[/U]
P(Green) = Total Green Marbles in the Jar / Total Marbles in the Jar
Total Green Marbles in the Jar = 8
Total Marbles in the Jar = 7 red + 8 green + 6 blue = 21
P(Green) = 8/21
[U]Draw #2:[/U]
P(Green) = Total Green Marbles in the Jar / Total Marbles in the Jar
Total Green Marbles in the Jar = 8 - 1 = 7
Total Marbles in the Jar = 7 red + 7 green + 6 blue = 20
P(Green) = 7/20
[U]Draw #3:[/U]
P(Green) = Total Green Marbles in the Jar / Total Marbles in the Jar
Total Green Marbles in the Jar = 7 - 1 = 6
Total Marbles in the Jar = 7 red + 6 green + 6 blue = 19
P(Green) = 6/19
[U]Draw #4:[/U]
P(Green) = Total Green Marbles in the Jar / Total Marbles in the Jar
Total Green Marbles in the Jar = 6 - 1 = 5
Total Marbles in the Jar = 7 red + 5 green + 6 blue = 18
P(Green) = 5/18
We want P(Green, Green, Green, Green)
Because each draw is [U][B]independent[/B][/U] of all other draws, we multiply each draw to get the final probability
P(Green, Green, Green, Green) = P(Green on Draw 1) * P(Green on Draw 2) * P(Green on Draw 3) * P(Green on Draw 4) *
P(Green, Green, Green, Green) = 8/21 * 7/20 * 6/19 * 5/18
P(Green, Green, Green, Green) = 1680/143640
Using our [URL='https://www.mathcelebrity.com/fraction.php?frac1=1680%2F143640&frac2=3%2F8&pl=Simplify']fraction simplifier[/URL], we get:
P(Green, Green, Green, Green) = [B]2/171
[MEDIA=youtube]b2C_D4_d0Ug[/MEDIA][/B]

a jar contains a $5 note, two $10 notes, a $20 note and a $50 note. if 2 notes are taken out by rand

a jar contains a $5 note, two $10 notes, a $20 note and a $50 note. if 2 notes are taken out by random, find the probability that their sum is $15
To get a sum of $15, we'd need to pull the $5 and the $10.
Since both events are indepdenent, we have:
P($5 or 10) or P(whatever is not pulled in the first pull)
First Pull: 2/4 (We can pull either a $10 or a $5, so 2 choices out of 4 bills)
Second Pull: 1/3 <-- since there are only 3 bills and 1 bill to pull
Each pull is independent, so we multiply:
2/4 * 1/3 = 2/12
We can simply this, so [URL='https://www.mathcelebrity.com/fraction.php?frac1=2%2F12&frac2=3%2F8&pl=Simplify']we type this fraction in our search engine[/URL] and we get:
[B]1/6[/B]

A laundry basket contains 30 socks, of which 9 are black. What is the probability that a randomly s

A laundry basket contains 30 socks, of which 9 are black. What is the probability that a randomly selected sock will be black?
P(Black) = 9/30
Simplifying, we can divide top and bottom by 3:
[B]3/10
3/10 as a percentage is 30%[/B]

A lottery uses a container with 25 identical balls numbered 1 through 25, from which three balls are

A lottery uses a container with 25 identical balls numbered 1 through 25, from which three balls are selected. What is the theoretical probability that the number 13 is picked first?
P(1st ball being 13) = [B]1 /25[/B]

A number cube is rolled and a coin is tossed. The number cube and the coin are fair. What is the pro

A number cube is rolled and a coin is tossed. The number cube and the coin are fair. What is the probability that the number rolled is greater than 3 and the coin toss is heads? Write your answer as a fraction in simplest form
Let's review the vitals of this question:
[LIST]
[*]The probability of heads on a fair coin is 1/2.
[*]On a fair die, greater than 3 means either 4, 5, or 6. Any die roll face is a 1/6 probability.
[*]So we have a combination of outcomes below:
[/LIST]
Outcomes
[LIST=1]
[*]Heads and 4
[*]Heads and 5
[*]Heads and 6
[/LIST]
For each of the outcomes, we assign a probability. Since the coin flip and die roll are independent, we multiply the probabilities:
[LIST=1]
[*]P(Heads and 4) = 1/2 * 1/6 = 1/12
[*]P(Heads and 5) = 1/2 * 1/6 = 1/12
[*]P(Heads and 6) = 1/2 * 1/6 = 1/12
[/LIST]
Since we want any of those events, we add all three probabilities
1/12 + 1/12 + 1/12 = 3/12
This fraction is not simplified. S[URL='https://www.mathcelebrity.com/fraction.php?frac1=3%2F12&frac2=3%2F8&pl=Simplify']o we type this fraction into our search engine, and choose Simplify[/URL].
We get a probability of [B]1/4[/B].
By the way, if you need a decimal answer or percentage answer instead of a fraction, we type in the following phrase into our search engine:
[URL='https://www.mathcelebrity.com/perc.php?num=1&den=4&pcheck=1&num1=+16&pct1=+80&pct2=+35&den1=+90&pct=+82&decimal=+65.236&astart=+12&aend=+20&wp1=20&wp2=30&pl=Calculate']1/4 to decimal[/URL]
Alternative Answers:
[LIST]
[*]For a decimal, we get [B]0.25[/B]
[*]For a percentage, we get [B]25%[/B]
[/LIST]

A pair of dice are rolled. Find the probability for P(Not 2 or Not 12).

A pair of dice are rolled. Find the probability for P(Not 2 or Not 12).
P(Not 2 or Not 12) = 1 - P(2) - P(12)
P(Not 2 or Not 12) = 1 - [URL='https://www.mathcelebrity.com/2dice.php?gl=1&pl=2&opdice=1&rolist=+2%2C3%2C9%2C10&dby=2%2C3%2C5&ndby=4%2C5&montect=+100']1/36[/URL] - [URL='https://www.mathcelebrity.com/2dice.php?gl=1&pl=12&opdice=1&rolist=+2%2C3%2C9%2C10&dby=2%2C3%2C5&ndby=4%2C5&montect=+100']1/36[/URL]
P(Not 2 or Not 12) = 34/36
[URL='https://www.mathcelebrity.com/fraction.php?frac1=34%2F36&frac2=3%2F8&pl=Simplify']P(Not 2 or Not 12)[/URL] = [B]17/18[/B]

A pair of dice is rolled. Find the probability of rolling a sum of not less than 5

A pair of dice is rolled. Find the probability of rolling a sum of not less than 5.
The phrase [I]not less than[/I] also means greater than or equal to. So we [URL='https://www.mathcelebrity.com/2dice.php?gl=3&pl=5&opdice=1&rolist=+&dby=&ndby=&montect=+']use our 2 dice calculator for a sum roll of 5 or greater[/URL] and we get:
[B]5/6[/B]

A Pairs of fair dice is tossed. What is the probability of not getting a sum 7 or 8?

A Pairs of fair dice is tossed. What is the probability of not getting a sum 7 or 8?
Not a 7 or 8 means 2 - 6 or 9 - 12
[URL='https://www.mathcelebrity.com/2dice.php?gl=5&pl=6&opdice=1&rolist=+&dby=&ndby=&montect=+']Using our 2-dice calculator[/URL], P(2 - 6) = 5/12
[URL='https://www.mathcelebrity.com/2dice.php?gl=3&pl=9&opdice=1&rolist=+&dby=&ndby=&montect=+']Using our 2-dice calculator[/URL], P(9 - 12) = 5/18
Since the sum could be either of these, we add probabilities:
P(Not a 7 or 8) = P(2 - 6) + P(9 - 12)
P(Not a 7 or 8) = 5/12 + 5/18
[URL='https://www.mathcelebrity.com/fraction.php?frac1=5%2F12&frac2=5%2F18&pl=Add']P(Not a 7 or 8) [/URL]= [B]25/36[/B]

A parking lot has seventy-one parking spaces numbered from 1 to 71. There are no cars in the parking

A parking lot has seventy-one parking spaces numbered from 1 to 71. There are no cars in the parking lot when Jillian pulls in and randomly parks. What is the probability that the number on the parking space where she parks is greater than or equal to 31?
Greater than or equal to means including 31 all the way through 71
31-71 is 40 spaces
P(s>=31) = [B]40/71[/B]

A parking lot has sixty-eight parking spaces numbered from 1 to 68. There are no cars in the parking

A parking lot has sixty-eight parking spaces numbered from 1 to 68. There are no cars in the parking lot when Jillian pulls in and randomly parks. What is the probability that the number on the parking space where she parks is greater than or equal to 21?
We want P(X>=21). This is also found by taking 1 - P(X <= 20).
P(X<=20) = 20/68. Reduced using a [URL='http://www.mathcelebrity.com/gcflcm.php?num1=20&num2=68&num3=&pl=GCF']GCF of 4[/URL], we get 5/17.
P(X >=21) = 1 - 5/17 = [B]12/17[/B]

A pet store has 10 rabbits. What’s the probability that they have at least 1 female rabbit.

A pet store has 10 rabbits. What’s the probability that they have at least 1 female rabbit.
At least 1 female rabbit means we [U]must[/U] have a female rabbit
First, we calculate the probability of 0 females
A rabbit can be either male or female with equal probabilities of 1/2 or 0.5.
Since each birth is independent, we can multiply to get the probability of all males:
P(MMMMMMMMMM) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2
P(MMMMMMMMMM) = 1/1024
Then, we subtract this probability from 1 to get the probability of [B]at least[/B] one female:
P(At least one F) = 1 - 1/1024
Since 1 = 1024/1024, we have:
P(At least one F) = (1024 - 1)/1024
P(At least one F) = [B]1023/1024[/B]

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leis

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.22 hours, with a standard deviation of 2.31 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.29 hours, with a standard deviation of 1.58 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children (u1 - u2)
What is the interpretation of this confidence interval?
A. There is 90% confidence that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours
B. There is a 90% probability that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours
C. There is 90% confidence that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours
D. There is a 90% probability that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours
0.2021 < u1 - u2 < 1.6579 using our [URL='http://www.mathcelebrity.com/meandiffconf.php?n1=+40&xbar1=+5.22&stdev1=2.31&n2=40&xbar2=4.29&stdev2=1.58&conf=+90&pl=Mean+Diff+Conf.+Interval+%28Large+Sample%29']difference of means confidence interval calculator[/URL]
[B]Choice D
There is a 90% probability that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours[/B]

A random variable X follows the uniform distribution with a lower limit of 670 and an upper limit

A random variable X follows the uniform distribution with a lower limit of 670 and an upper limit
a. Calculate the mean and standard deviation of this distribution. (Round intermediate calculation for standard deviation to 4 decimal places and final answer to 2 decimal places.)
Using our [URL='http://www.mathcelebrity.com/uniform.php?a=+670&b=+770&x=+680&t=+3&pl=PDF']uniform distribution calculator[/URL], we get:
[B]Mean = 720
Standard deviation = 28.87
[/B]
b. What is the probability that X is less than 730? (Do not round intermediate calculations. Round your answer to 2 decimal places.)
Using our [URL='http://www.mathcelebrity.com/uniform.php?a=+670&b=+770&x=+730&t=+3&pl=CDF']uniform distribution calculator[/URL], we get:
[B]0.6[/B]

A spinner has 10 equally sized sections, 2 are gray 8 are blue. The spinner is spun twice. What is t

A spinner has 10 equally sized sections, 2 are gray 8 are blue. The spinner is spun twice. What is the probability that the first spin lands on blue and the second lands on gray?
P(blue) = Blue sections / Total Sections
P(blue) = 8/10
[URL='https://www.mathcelebrity.com/fraction.php?frac1=8%2F10&frac2=3%2F8&pl=Simplify']Reducing this using our simplified fraction calculator[/URL], we get:
P(blue) = 4/5
P(gray) = Gray sections / Total Sections
P(blue) = 2/10
[URL='https://www.mathcelebrity.com/fraction.php?frac1=2%2F10&frac2=3%2F8&pl=Simplify']Reducing this using our simplified fraction calculator[/URL], we get:
P(gray) = 1/5
We want the probability of blue,gray. Since each spin is independent, we multiply the two probabilities to get our answer:
P(blue, gray) = P(blue) * P(gray)
P(blue, gray) = 4/5 * 1/5
P(blue, gray) = [B]4/25[/B]

A spinner has 3 equal sections labelled A, B, C. A bag contains 3 marbles: 1 grey, 1 black, and 1 w

A spinner has 3 equal sections labelled A, B, C. A bag contains 3 marbles: 1 grey, 1 black, and 1 white. The pointer is spun and a marble is picked at random.
a) Use a tree diagram to list the possible outcomes.
[LIST=1]
[*][B]A, Grey[/B]
[*][B]A, Black[/B]
[*][B]A, White[/B]
[*][B]B, Grey[/B]
[*][B]B, Black[/B]
[*][B]B, White[/B]
[*][B]C, Grey[/B]
[*][B]C, Black[/B]
[*][B]C, White[/B]
[/LIST]
b) What is the probability of:
i) spinning A?
P(A) = Number of A sections on spinner / Total Sections
P(A) = [B]1/3[/B]
---------------------------------
ii) picking a grey marble?
P(A) = Number of grey marbles / Total Marbles
P(A) = [B]1/3[/B]
---------------------------------
iii) spinning A and picking a white marble?
Since they're independent events, we multiply to get:
P(A AND White) = P(A) * P(White)
P(A) was found in i) as 1/3
Find P(White):
P(White) = Number of white marbles / Total Marbles
P(White) = 1/3
[B][/B]
Therefore, we have:
P(A AND White) = 1/3 * 1/3
P(A AND White) = [B]1/9[/B]
---------------------------------
iv) spinning C and picking a pink marble?
Since they're independent events, we multiply to get:
P(C AND Pink) = P(C) * P(Pink)
Find P(C):
P(C) = Number of C sections on spinner / Total Sections
P(C) = 1/3
[B][/B]
Find P(Pink):
P(Pink) = Number of pink marbles / Total Marbles
P(Pink) = 0/3
[B][/B]
Therefore, we have:
P(C AND Pink) = 1/3 * 0
P(C AND Pink) = [B]0[/B]

A spinner has 6 equal sections, of which 2 are green. If you spin the spinner once, what is the prob

A spinner has 6 equal sections, of which 2 are green. If you spin the spinner once, what is the probability that it will land on a green section? Write your answer as a fraction or whole number.
P(green) = Total Green / Total spaces
P(green) = 2/6
We can simplify this fraction. So we [URL='https://www.mathcelebrity.com/fraction.php?frac1=2%2F6&frac2=3%2F8&pl=Simplify']type 2/6 into our search engine[/URL], choose Simplify, and we get:
P(green) = [B]1/3[/B]

A spinner is divided into 4 equal sections numbered 1 to 4. The theoretical probability of the spinn

A spinner is divided into 4 equal sections numbered 1 to 4. The theoretical probability of the spinner stopping on 3 is 25%. Which of the following is most likely the number of 3s spun in 10,000 spins?
We want Expected Value of s spins. Set up the expected value formula for any number 1-4
E(s) = 0.25 * n where n is the number of spins.
Using s = 3, n = 10,000, we have:
E(10,000) = 0.25 * 10,000
E(10,000) = [B]2,500[/B]

A standard die is rolled. Find the probability that the number rolled is greater than 3

A standard die is rolled. Find the probability that the number rolled is greater than 3.
Using our [URL='http://www.mathcelebrity.com/1dice.php?gl=2&pl=3&opdice=1&rolist=+2%2C3%2C4&dby=+2%2C3%2C5&montect=+100']dice calculator[/URL], the probability is [B]1/2 or 0.5[/B]

A student hypothesized that girls in his class had the same blood pressure levels as boys. The proba

A student hypothesized that girls in his class had the same blood pressure levels as boys. The probability value for his null hypothesis was 0.15. So he concluded that the blood pressures of the girls were higher than boys'. Which kind of mistake did he make?
a. Type I error
b. Type II error
c. Type I and Type II error
d. He did not make any mistake
[B]d. He did not make any mistake[/B]
[I]p value is high, especially using a significance level of 0.05[/I]

A teacher hypothesized that in her class, grades of girls on a chemistry test were the same as grade

A teacher hypothesized that in her class, grades of girls on a chemistry test were the same as grades of boys. If the probability value of her null hypothesis was 0.56, it suggested:
a. We failed to reject the null hypothesis
b. Boys' grades were higher than girls' grades
c. Girls' grades were higher than boys' grades
d. The null hypothesis was rejected
[B]a. We failed to reject the null hypothesis[/B]
Due to a high probability.

A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selec

A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selected adults are given an IQ test, what is the probability that the sample mean scores will be between 85 and 125 points?
For x = 125, our z-score and probability is seen [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+125&mean=+105&stdev=+20&n=+1&pl=P%28X+%3C+Z%29']here[/URL]
Z = 1
P(x < 1) = 0.841345
For x = 85, our z-score and probability is seen [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+85&mean=+105&stdev=+20&n=+1&pl=P%28X+%3C+Z%29']here[/URL]
Z = -1
P(x < -1) = 0.158655
So what we want is the probability between these values:

0.841345 - 0.158655 = [B]0.68269[/B]

0.841345 - 0.158655 = [B]0.68269[/B]

Aaron buys a bag of cookies that contains 8 chocolate chip cookies, 6 peanut butter cookies,7 sugar

Aaron buys a bag of cookies that contains 8 chocolate chip cookies, 6 peanut butter cookies,7 sugar cookies and 6 oatmeal raisin cookies. What it’s the probability that Aaron randomly selects a peanut butter cookie from the bag, eats it,, then randomly selects another peanut butter cookie?
First draw out of the bag is a peanut butter cookie:
P(PB) = Total Peanut Butter Cookies / Total Cookies
P(PB) = 6/27
Second draw out of the bag is a peanut butter cookie, but we have one less since Aaron ate one:
P(PB) = Total Peanut Butter Cookies - 1 / Total Cookies - 1
P(PB) = (6 - 1)/(27 - 1)
P(PB) = 5/26
Now, since each event is independent, we multiply them to see the probability of choosing a peanut butter cookie, eating it, then reaching in and choosing another peanut butter cookie:
P(PB, PB) = 6/27 * 5/26
[URL='https://www.mathcelebrity.com/fraction.php?frac1=6%2F27&frac2=5%2F26&pl=Multiply']P(PB, PB)[/URL] = [B]5/117[/B]

An analysis of the final test scores for Managerial Decision Making Tools reveals the scores follow

An analysis of the final test scores for Managerial Decision Making Tools reveals the scores follow the normal probability distribution. The mean of the distribution is 75 and the standard deviation is 8. The instructor wants to award an "A" to students whose score is in the highest 10 percent. What is the dividing point for those students who earn an "A"?
Top 10% is equivalent to the 90th percentile.
Using our [URL='http://www.mathcelebrity.com/percentile_normal.php?mean=+75&stdev=8&p=+90&pl=Calculate+Percentile']percentile calculator[/URL], the 90th percentile cutoff point is [B]85.256[/B]

An eccentric millionaire decided to give away $1,000,000 if Janelle took one die and rolled a "4". H

An eccentric millionaire decided to give away $1,000,000 if Janelle took one die and rolled a "4". He wanted Janelle to have a better than 1 in 6 chance of winning, so before she rolled the die he told her that she could roll the die 3 times. If any roll was a "4", she would win the million dollars. What are Janelle's chances of winning the million dollars?
Chance of winning each roll is 1/6. Which means the chances of losing each roll are 1 - 1/6 = 5/6
Calculate the probability of 3 straight losing rolls:
P(Lose) = P(Loser) * P(Loser) * P(Loser) = 5/6 * 5/6 * 5/6 = 125/216
P(Win) = 1 - P(Lose)
P(Win) = 1 - 125/216
P(Win) = 216/216 -125/216
P(Win) = [B]91/216[/B]

An ordinary fair die is rolled twice. The face value of the rolls is added together. Compute the pro

An ordinary fair die is rolled twice. The face value of the rolls is added together. Compute the probability of the following events: Event A: The sum is greater than 6. Event B: The sum is divisible by 5 or 6 or both.
[URL='http://www.mathcelebrity.com/2dice.php?gl=2&pl=6&opdice=1&rolist=+2%2C3%2C9%2C10&dby=2%2C3%2C5&ndby=4%2C5&montect=+100']Sum greater than 6[/URL] = [B]7/12[/B]
Sum is divisible by 5 or 6 or both
This means a sum of 5, a sum of 6, a sum of 10, or a sum of 12.
[URL='http://www.mathcelebrity.com/2dice.php?gl=1&pl=5&opdice=1&rolist=+2%2C3%2C9%2C10&dby=2%2C3%2C5&ndby=4%2C5&montect=+100']Sum of 5[/URL] = 1/9 or 4/36
[URL='http://www.mathcelebrity.com/2dice.php?gl=1&pl=6&opdice=1&rolist=+2%2C3%2C9%2C10&dby=2%2C3%2C5&ndby=4%2C5&montect=+100']Sum of 6[/URL] = 5/36
[URL='http://www.mathcelebrity.com/2dice.php?gl=1&pl=10&opdice=1&rolist=+2%2C3%2C9%2C10&dby=2%2C3%2C5&ndby=4%2C5&montect=+100']Sum of 10[/URL] = 1/12 or 3/36
[URL='http://www.mathcelebrity.com/2dice.php?gl=1&pl=12&opdice=1&rolist=+2%2C3%2C9%2C10&dby=2%2C3%2C5&ndby=4%2C5&montect=+100']Sum of 12[/URL] = 1/36
Adding all these up, we get:
(4 + 5 + 3 + 1)/36
[B]13/36[/B]

As the sample size increases, we assume:

As the sample size increases, we assume:
a. α increases
b. β increases
c. The probability of rejecting a hypothesis increases
d. Power increases
[B]d. Power increases[/B]
[LIST]
[*]Power increases if the standard deviation is smaller.
[*]If the difference between the means is bigger, the power is bigger.
[*]Sample size also increases power
[/LIST]

Assume that you make random guesses for 5 true-or-false questions

Assume that you make random guesses for 5 true-or-false questions.
(a) What is the probability that you get all 5 answers correct? (Show work and write the answer in simplest fraction form)
(b) What is the probability of getting the correct answer in the 5th question, given that the first four answers are all wrong? (Show work and write the answer in simplest fraction form)
(c) If event A is “Getting the correct answer in the 5th question” and event B is “The first four answers are all wrong”. Are event A and event B independent? Please explain.
(a) Correct Answer on each one is 1/2 or 0.5. Since all are independent events, we have:
(1/2)^5 = [B]1/32[/B]
(b) We have [B]1/2[/B]
(1/2)^4 * 1/2/((1/2)^4)
c) [B]Independent since you could have gotten correct or wrong on any of the 4 and the probability does not change[/B]

Assume you have a laptop worth 2900. There is a 3 percent chance of it getting lost what’s the fair

Assume you have a laptop worth 2900. There is a 3 percent chance of it getting lost what’s the fair premium insurance?
Fair premium Insurance = Price * probability of loss:
Fair premium Insurance = 2,900 * 3%
Fair premium Insurance = [B]87[/B]

assume your math class has 10 sophomores and 7 juniors. there are 3 female sophomores and 4 male jun

assume your math class has 10 sophomores and 7 juniors. there are 3 female sophomores and 4 male juniors. what is the probability of randomly selecting a student who is a female or a junior
[U]Sophomores:[/U]
10 sophomores:
3 female
male = 10 - 3 = 7
[U]Juniors:[/U]
7 juniors
4 males
female = 7 - 4 = 3
[U]Females:[/U]
Total Females = Female Sophomores + Female Juniors
Total Females = 3 + 3
Total Females = 6
[U]Total Students:[/U]
Total Students = Total Sophomores + Total Juniors
Total Students = 10 + 7
Total Students = 17
[U]We want P(female or Junior). We use the formula below to avoid duplicates:[/U]
P(female or Junior) = P(female) + P(junior) - P(female and junior)
P(female or Junior) = Total Females / Total Students + Total Juniors / Total Students - Total Female Juniors / Total Students
P(female or Junior) = 6/17 + 7/17 - 3/17
P(female or Junior) = [B]10/17[/B]

Assuming a standard 52-card deck, what's the probability of dealing three eights in a row when the c

Assuming a standard 52-card deck, what's the probability of dealing three eights in a row when the cards are returned and the deck is shuffled between each draw?
There are four (8's) in a standard 52 card deck. The probability of drawing an 8 is:
4/52
[URL='https://www.mathcelebrity.com/fraction.php?frac1=4%2F52&frac2=3%2F8&pl=Simplify']Using our fraction simplifier[/URL], we get:
1/13
Now, with each draw, we replace the deck. So each draw of an 8 has a 1/13 probability. And since each of the three draws is independent, we multiply each probability:
1/13 * 1/13 * 1/13 = [B]1/2197 or 0.00045516613[/B]

At a certain university, 60% of the students enrolled in a math course, 50% are enrolled in an Engli

At a certain university, 60% of the students enrolled in a math course, 50% are enrolled in an English course, and 40% are enrolled in both. What percentage of the students are enrolled in an English course and/or a math course?
Let M be a math course, E be an english course, We are given:
[LIST]
[*]P(M) = 0.6
[*]P(E) = 0.5
[*]P(E AND M) = 0.4
[*]We want P(E U M)
[/LIST]
Using [URL='http://www.mathcelebrity.com/probunion2.php?pa=0.6+&pb=+0.5&paintb=+0.4&aub=+&pl=Calculate']two event probability[/URL], we get [B]P(E U M) = 0.7[/B]

Balls numbered 1 to 10 are placed in a bag. Two of the balls are drawn out at random. Find the proba

Balls numbered 1 to 10 are placed in a bag. Two of the balls are drawn out at random. Find the probability that the numbers on the balls are consecutive.
Build our sample set:
[LIST]
[*](1, 2)
[*](2, 3)
[*](3, 4)
[*](4, 5)
[*](5, 6)
[*](6, 7)
[*](7, 8)
[*](8, 9)
[*](9, 10)
[/LIST]
Each of these 9 possibilities has a probability of:
1/10 * 1/9
This is because we draw without replacement. To start, the bag has 10 balls. On the second draw, it only has 9. We multiply each event because each draw is independent.
We have 9 possibilities, so we have:
9 * 1/10 * 1/9
Cancelling, the 9's, we have [B]1/10[/B]

Basic Statistics

Free Basic Statistics Calculator - Given a number set, and an optional probability set, this calculates the following statistical items:

Expected Value

Mean = μ

Variance = σ^{2}

Standard Deviation = σ

Standard Error of the Mean

Skewness

Mid-Range

Average Deviation (Mean Absolute Deviation)

Median

Mode

Range

Pearsons Skewness Coefficients

Entropy

Upper Quartile (hinge) (75th Percentile)

Lower Quartile (hinge) (25th Percentile)

InnerQuartile Range

Inner Fences (Lower Inner Fence and Upper Inner Fence)

Outer Fences (Lower Outer Fence and Upper Outer Fence)

Suspect Outliers

Highly Suspect Outliers

Stem and Leaf Plot

Ranked Data Set

Central Tendency Items such as Harmonic Mean and Geometric Mean and Mid-Range

Root Mean Square

Weighted Average (Weighted Mean)

Frequency Distribution

Successive Ratio

Expected Value

Mean = μ

Variance = σ

Standard Deviation = σ

Standard Error of the Mean

Skewness

Mid-Range

Average Deviation (Mean Absolute Deviation)

Median

Mode

Range

Pearsons Skewness Coefficients

Entropy

Upper Quartile (hinge) (75th Percentile)

Lower Quartile (hinge) (25th Percentile)

InnerQuartile Range

Inner Fences (Lower Inner Fence and Upper Inner Fence)

Outer Fences (Lower Outer Fence and Upper Outer Fence)

Suspect Outliers

Highly Suspect Outliers

Stem and Leaf Plot

Ranked Data Set

Central Tendency Items such as Harmonic Mean and Geometric Mean and Mid-Range

Root Mean Square

Weighted Average (Weighted Mean)

Frequency Distribution

Successive Ratio

Bayes Rule

Free Bayes Rule Calculator - Calculates the conditional probabilities of (B given A) of 2 events and a conditional probability event using Bayes Rule

Bernoulli Trials

Free Bernoulli Trials Calculator - Given a success probability p and a number of trials (n), this will simulate Bernoulli Trials and offer analysis using the Bernoulli Distribution. Also calculates the skewness, kurtosis, and entropy

Binomial Distribution

Free Binomial Distribution Calculator - Calculates the probability of 3 separate events that follow a binomial distribution. It calculates the probability of exactly k successes, no more than k successes, and greater than k successes as well as the mean, variance, standard deviation, skewness and kurtosis.

Also calculates the normal approximation to the binomial distribution with and without the continuity correction factor

Calculates moment number t using the moment generating function

Also calculates the normal approximation to the binomial distribution with and without the continuity correction factor

Calculates moment number t using the moment generating function

Binominal Probability

If a seed is planted, it has a 90% chance of growing into a healthy plant.
If 12 seeds are planted, what is the probability that exactly 4 don't grow?
Im seriously confused is it like u multiple the amount of the (0.90) and multiple (0.30) by power depends how any they r right?

Binominal Probability

You want the binomial distribution, where a "success" is that the plant [U]does not[/U] grow.
So if the probability that the plant grows is 0.9, the probability it does not grow is 1 - 0.9 = 0.1.
We have n = 12, p = 0.1
You want the probability that exactly 4 of 12 do not grow.
Use our [URL='http://www.mathcelebrity.com/binomial.php?n=+12&p=+0.1&k=+4&t=+5&pl=P%28X+%3D+k%29']binomial distribution probability calculato[/URL]r to get P(X = 4) = [B]0.0213[/B]

can 0.2 be the probability of an outcome in a sample space?

can 0.2 be the probability of an outcome in a sample space?
Yes.
Any probability p is a valid sample space outcome if:
[B]0 <= p <= 1[/B]

Chebyshevs Theorem

Free Chebyshevs Theorem Calculator - Using Chebyshevs Theorem, this calculates the following:

Probability that random variable X is within k standard deviations of the mean.

How many k standard deviations within the mean given a P(X) value.

Probability that random variable X is within k standard deviations of the mean.

How many k standard deviations within the mean given a P(X) value.

CHEBYSHEVS THEOREM TELLS US THAT WHAT PERCENTAGE LIES BETWEEN 2.25 STANDARD DEVIATIONS?

CHEBYSHEVS THEOREM TELLS US THAT WHAT PERCENTAGE LIES BETWEEN 2.25 STANDARD DEVIATIONS?
Using our [URL='http://www.mathcelebrity.com/chebyshev.php?pl=probability&k=2.25&probk=0.75']Chebyshevs Theorem calculator[/URL], we get:
P(X - u| < kσ) >= [B]0.802469[/B]

Chi-Square Critical Values

Free Chi-Square Critical Values Calculator - Given a probability, this calculates the critical value for the right-tailed and left-tailed tests for the Chi-Square Distribution. CHIINV from Excel is used as well.

Chuck-a-luck is an old game, played mostly in carnivals and county fairs. To play chuck-a-luck you p

Chuck-a-luck is an old game, played mostly in carnivals and county fairs. To play chuck-a-luck you place a bet, say $1, on one of the numbers 1 through 6. Say that you bet on the number 4. You then roll three dice (presumably honest). If you roll three 4’s, you win $3.00; If you roll just two 4’s, you win $2; if you roll just one 4, you win $1 (and, in all of these cases you get your original $1 back). If you roll no 4’s, you lose your $1. Compute the expected payoff for chuck-a-luck.
Expected payoff for each event = Event Probability * Event Payoff
Expected payoff for 3 matches:
3(1/6 * 1/6 * 1/6) = 3/216 = 1/72
Expected payoff for 2 matches:
2(1/6 * 1/6 * 5/6) = 10/216 = 5/108
Expected payoff for 1 match:
1(1/6 * 5/6 * 5/6) = 25/216
Expected payoff for 0 matches:
-1(5/6 * 5/6 * 5/6) = 125/216
Add all these up:
(3 + 10 + 25 - 125)/216
-87/216 ~ [B]-0.40[/B]

Coin Toss Probability

Free Coin Toss Probability Calculator - This calculator determines the following coin toss probability scenarios

* Coin Toss Sequence such as HTHHT

* Probability of x heads and y tails

* Probability of at least x heads in y coin tosses

* Probability of at least x tails in y coin tosses

* Probability of no more than x heads in y coin tosses

* Probability of no more than x tails in y coin tosses

* (n) Coin Tosses with a list of scenario results displayed

* Monte Carlo coin toss simulation

* Coin Toss Sequence such as HTHHT

* Probability of x heads and y tails

* Probability of at least x heads in y coin tosses

* Probability of at least x tails in y coin tosses

* Probability of no more than x heads in y coin tosses

* Probability of no more than x tails in y coin tosses

* (n) Coin Tosses with a list of scenario results displayed

* Monte Carlo coin toss simulation

Consider a firm that has two assembly lines, 1 and 2, both producing calculator. Assume that you hav

Consider a firm that has two assembly lines, 1 and 2, both producing calculator. Assume that you have purchased a calculator and it turns out to be defective. And the line 1 produces 60% of all calculators produced.
L1: event that the calculator is produced on line 1
L2: event that the calculator is produced on line 2
Suppose that your are given the conditional information:
10% of the calculator produced on line 1 is defective
20% of the calculator produced on line 2 is defective
Q: If we choose one defective, what is the probability that the defective calculator comes from Line 1 and Line2?
L1 = event that the calculator is produced on line 1 = 0.6
L2 = event that the calculator is produced on line 2 = 1 - 0.6 = 0.4
D = Defective
D|L1 Defective from Line 1 = 0.1
D|L2 = Defective from Line 2 = 0.20
[U]Defective from Line 1[/U]
P(L1|D) = P(L1)P(D/L1) / [ P(L1)P(D/L1) + P(L2)P(D/L2)]
P(L1|D) = (.60)(.10) /[(.60)(.10)+ (.40)(.20)]
[B]P(L1|D) = 0.4286[/B]
[U]Defective from Line 2[/U]
P(L2|D) = P(L2)P(D/L2) / [ P(L1)P(D/L1) + P(L2)P(D/L2)]
P(L2|D) = (.40)(.20) /[(.60)(.10)+ (.40)(.20)]
[B]P(L2|D) = 0.5714[/B]

Consider a probability model consisting of randomly drawing two colored balls from a jar containing

Consider a probability model consisting of randomly drawing two colored balls from a jar containing 2 red and 1 blue balls. What is the Sample Space of this experiment? (assume B= blue and R=red)
The sample space is the list of all possible events
[LIST]
[*]RRB
[*]RBR
[*]BRR
[/LIST]

Consider the case of a manufacturer who has an automatic machine that produces an important part. Pa

Consider the case of a manufacturer who has an automatic machine that produces an important part. Past records indicate that at the beginning of the data the machine is set up correctly 70 percent of the time. Past experience also shows that if the machine is set up correctly it will produce good parts 90 percent of the time. If it is set up incorrectly, it will produce good parts 40 percent of the time. Since the machine will produce 60 percent bad parts, the manufacturer is considering using a testing procedure. If the machine is set up and produces a good part, what is the revised probability that it is set up correctly?
[U]Determine our events:[/U]
[LIST]
[*]C = Correctly Set Machine = 0.7
[*]C|G = Correctly Set Machine And Good Part = 0.9
[*]I = Incorrectly Set Machine = 1 - 0.7 = 0.3
[*]I|G = Incorrectly Set Machine And Good Part = 0.4
[*]B< = BAD PARTS = 0.60
[/LIST]
P[correctly set & part ok] = P(C) * P(C|G)
P[correctly set & part ok] = 70% * 90% = 63%
P[correctly set & part ok] = P(I) * P(I|G)
P[incorrectly set & part ok] = 30% *40% = 12%
P[correctly set | part ok] = P[correctly set & part ok]/(P[correctly set & part ok] + P[incorrectly set & part ok])
P[correctly set | part ok] = 63/(63+12) = [B]0.84 or 84%[/B]

Container Arrangements

Free Container Arrangements Calculator - Given a set of items inside a container, this calculates the probability that you draw certain items in the following fashion:

Draw__all__ the items

Draw__any of__ the items

How many ways can you choose m items of a, n items of b, o items of c, etc.

Draw

Draw

How many ways can you choose m items of a, n items of b, o items of c, etc.

Conventionally, the null hypothesis is false if the probability value is: a. Greater than 0.05 b. L

Conventionally, the null hypothesis is false if the probability value is:
a. Greater than 0.05
b. Less than 0.05
c. Greater than 0.95
d. Less than 0.95
[B]b. Less than 0.05[/B]
This is standard in hypothesis testing using p = 0.05

Critical Z-values

Free Critical Z-values Calculator - Given a probability from a normal distribution, this will generate the z-score critical value. Uses the NORMSINV Excel function.

Customers arrive at the claims counter at the rate of 20 per hour (Poisson distributed). What is th

Customers arrive at the claims counter at the rate of 20 per hour (Poisson distributed). What is the probability that the arrival time between consecutive customers is less than five minutes?
Use the [I]exponential distribution[/I]
20 per 60 minutes is 1 every 3 minutes
1/λ = 3 so λ = 0.333333333
Using the [URL='http://www.mathcelebrity.com/expodist.php?x=+5&l=0.333333333&pl=CDF']exponential distribution calculator[/URL], we get F(5,0.333333333) = [B]0.811124396848[/B]

Dane wrote the letters of “NEW YORK CITY” on cards and placed them in a hat. What is the probability

Dane wrote the letters of “NEW YORK CITY” on cards and placed them in a hat. What is the probability that he will draw the letter “Y” out of the hat?
New York City has 11 letters.
Our probability of drawing a Y is denoted as P(Y):
P(Y) = Number of Y's / Total Letters
P(Y) = [B]2/11[/B]

Determine the area under the standard normal curve that lies between:

Determine the area under the standard normal curve that lies between:
(a) Z = -0.38 and Z = 0.38
(b) Z = -2.66 and Z = 0
(c) Z = -1.04 and Z - 1.67
[B](a) 0.2961 using our [URL='http://www.mathcelebrity.comzscore.php?z=+p%28-0.38%3Cz%3C0.38%29&pl=Calculate+Probability']z score calculator[/URL]
(b) 0.4961 using our [URL='http://www.mathcelebrity.com/zscore.php?z=+p%28-2.66%3Cz%3C0%29&pl=Calculate+Probability']z score calculator[/URL]
(c) 0.8034 using our [URL='http://www.mathcelebrity.com/zscore.php?z=+p%28-1.04%3Cz%3C1.67%29&pl=Calculate+Probability']z score calculator[/URL][/B]

Each of letters in the word PROPER are on separate cards, face down on the table. If you pick a card

Each of letters in the word PROPER are on separate cards, face down on the table. If you pick a card at random, what is the probability that its letter will be P or R?
PROPER has 6 letters in it. It has 2 P's and 2 R's. So we have:
Pr(P or R) = Pr(P) + Pr(R)
Pr(P or R) = 2/6 + 2/6
Pr(P or R) = 4/6
We can simplify this. So [URL='https://www.mathcelebrity.com/fraction.php?frac1=4%2F6&frac2=3%2F8&pl=Simplify']we type this fraction in our search engine[/URL], choose simplify, and we get:
Pr(P or R) = [B]2/3[/B]

Each of the letters in the word PLOTTING are on separate cards, face down on the table. If you pick

Each of the letters in the word PLOTTING are on separate cards, face down on the table. If you pick a card at random, what is the probability that its letter will be T or G?
PLOTTING has to 8 letters. It has 2 T'sand 1 G, so we have:
P(T or G) = P(T) + P(G)
P(T or G) = 2/8 + 1/8
P(T or G) = [B]3/8[/B]

Eighty percent of the employees at Rowan University have their biweekly Wages deposit directly to th

Eighty percent of the employees at Rowan University have their biweekly Wages deposit directly to their bank by electronic deposit program. Suppose we select a random samples of 8 employees. What is the probability that three of the eight (8) sampled employees use direct deposit program?
Use the [I]binomial distribution[/I]
[LIST]
[*]p = 0.8
[*]n = 8
[*]k = 3
[/LIST]
So we want P(X = 3)
Using our [URL='http://www.mathcelebrity.com/binomial.php?n=+8&p=+0.8&k=+3&t=+5&pl=P%28X+=+k%29']binomial distribution calculator[/URL], we get P(X = 3) = [B]0.0092[/B]

Erik is rolling two regular six-sided number cubes. What is the probability that he will roll an eve

Erik is rolling two regular six-sided number cubes. What is the probability that he will roll an even number on one cube and a prime number on the other?
P(Even on first cube) = (2,4,6) / 6 total choices
P(Even on first cube) = 3/6
P(Even on first cube) = 1/2 <-- [URL='https://www.mathcelebrity.com/fraction.php?frac1=3%2F6&frac2=3%2F8&pl=Simplify']Using our fraction simplify calculator[/URL]
P(Prime on second cube) = (2,3,5) / 6 total choices
P(Prime on second cube) = 3/6
P(Prime on second cube) = 1/2 <-- [URL='https://www.mathcelebrity.com/fraction.php?frac1=3%2F6&frac2=3%2F8&pl=Simplify']Using our fraction simplify calculator[/URL]
Since each event is independent, we have:
P(Even on the first cube, Prime on the second cube) = P(Even on the first cube) * P(Prime on the second cube)
P(Even on the first cube, Prime on the second cube) = 1/2 * 1/2
P(Even on the first cube, Prime on the second cube) = [B]1/4[/B]

Event Likelihood

Free Event Likelihood Calculator - Given a probability, this determines how likely that event is

Exponential Distribution

Free Exponential Distribution Calculator - Calculates the Probability Density Function (PDF) and Cumulative Density Function (CDF) of the exponential distribution as well as the mean, variance, standard deviation, and entropy.

Facebook provides a variety of statistics on its Web site that detail the growth and popularity of t

Facebook provides a variety of statistics on its Web site that detail the growth and popularity of the site.
On average, 28 percent of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent.
a. Find the probability that the percent of 18 to 34-year-olds who check Facebook before getting out of bed in the morning is at least 30.
b. Find the 95th percentile, and express it in a sentence.
a. P(X >=0.30), calculate the [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+0.30&mean=+0.28&stdev=+0.05&n=+1&pl=P%28X+%3E+Z%29']z-score[/URL] which is:
Z = 0.4
P(x>0.4) = [B]0.344578 or 34.46%[/B]
b. Inverse Normal (0.95) [URL='http://www.mathcelebrity.com/zcritical.php?a=0.95&pl=Calculate+Critical+Z+Value']calculator[/URL] = 1.644853627
Use NORMSINV(0.95) on Excel
0.28 + 0.05(1.644853627) = [B]0.362242681 or 36.22%[/B]

Farah rolls a fair dice and flips a fair coin. What is the probability of obtaining a 5 and a head?

Farah rolls a fair dice and flips a fair coin. What is the probability of obtaining a 5 and a head? Give your answer in its simplest form.
Probability of a 5 is 1/6
Probability of a head is 1/2
Since each event is independent, we get the total probability by multiplying both together:
P(5,H) = 1/6 * 1/2
P(5,H) = [B]1/12[/B]

Favorable Outcome

Free Favorable Outcome Calculator - Shows you various examples of favorable outcome in probability

find the probability of drawing a 4 or an ace

find the probability of drawing a 4 or an ace.
In a 52 card deck, there are 4 (4's) and 4 (Aces), for a total of 8 cards:
The probability is 8 cards / 52 total cards:
8/52 using our s[URL='https://www.mathcelebrity.com/fraction.php?frac1=8%2F52&frac2=3%2F8&pl=Simplify']implify fractions calculator[/URL] = [B]2/13[/B]

Fishers Exact Test

Free Fishers Exact Test Calculator - Given a, b, c, and d, this calculates the probability of any such set of values using Fishers exact Test

For a population with μ = 60 and σ = 12, what is the z-score that corresponds to a score of 66?

For a population with μ = 60 and σ = 12, what is the z-score that corresponds to a score of 66?
[URL='https://www.mathcelebrity.com/probnormdist.php?xone=66&mean=60&stdev=12&n=1&pl=P%28X+%3C+Z%29']Using our z-score calculator[/URL], we get a probability:
[B]0.691462[/B]

Four coins are flipped. What is the probability of the coins all landing on heads

Four coins are flipped. What is the probability of the coins all landing on heads
The probability of one head is 1/2. Since all 4 flips are independent, we multiply each flip probability:
P(HHHH) = 1/2 * 1/2 * 1/2 * 1/2
P(HHHH) = [B]1/16[/B]

From a regular deck of 52 playing cards, you turn over a 6 and then a 7. What is the probability tha

From a regular deck of 52 playing cards, you turn over a 6 and then a 7. What is the probability that the next card you turn over will be a face card?
Key phrases: 52 card standard deck so you know there's no tricks or missing cards.
[U]Calculate the number of face cards in a standard 52 card deck[/U]
First, we know that face cards = (J, K, Q)
We also know that there are 4 suits (Hearts, Diamonds, Spades, Clubs)
Total Face Cards = 3 face card types * 4 possible suits = 12 face cards
[U]Calculate total face down cards[/U]
First card, you turn over a 6
Next card, you turn over a 7
This means, we have 52 cards - 2 cards from the draws = 50 cards left in the deck which are face down.
P(Face Card) = Total Face Cards / Total Cards in the Deck Face Down
P(Face Card) = 12/50
Simplifying this fraction [URL='https://www.mathcelebrity.com/fraction.php?frac1=12%2F50&frac2=3%2F8&pl=Simplify']using our math engine[/URL], we get:
P(Face Card) = [B]6/25[/B]

Geometric Distribution

Free Geometric Distribution Calculator - Using a geometric distribution, it calculates the probability of exactly k successes, no more than k successes, and greater than k successes as well as the mean, variance, standard deviation, skewness, and kurtosis.

Calculates moment number t using the moment generating function

Calculates moment number t using the moment generating function

Given that P (A)=0.6, P (B)=0.5, P (A|B) = 0.2, P (C|A)= 0.3 and P (C|B)=0.4. (1) If they are depe

Given that P (A)=0.6, P (B)=0.5, P (A|B) = 0.2, P (C|A)= 0.3 and P (C|B)=0.4.
(1) If they are dependent each other, what is P (B | A) = ?
(2) If the event C is conditionally dependent upon evens A and B, What's the probability: P (A|C) = ?
(1) Bayes Rule: P(B|A) = P(B) * P(A|B)
P(B|A) = 0.5 * 0.2 = 0.1
(2) Bayes Rule: P(A|C) = P(A) * P(C|A)
P(A|C)= 0.6 * 0.3 = 0.18

Gregg has 8 cards.Half red,half black. He picks 2 cards from the deck.What is the probability both o

Gregg has 8 cards.Half red,half black. He picks 2 cards from the deck.What is the probability both of them are red?
Half means 4 cards are red and 4 cards are black.
The first draw probability of red is:
4 total red cards out of 8 total cards = 4/8.
[URL='https://www.mathcelebrity.com/fraction.php?frac1=4%2F8&frac2=3%2F8&pl=Simplify']Simplified, this is[/URL] 1/2
The second draw is 3 total red cards out of 7 remaining cards. Since 1 red was drawn (4 - 1) = 3 reds left and 1 card was drawn (8 -1) = left
3/7
Since each draw is independent, we multiply the probabilities:
1/2 * 3/7 = [B]3/14[/B]

Hardy-Weinberg

Free Hardy-Weinberg Calculator - Given a dominant gene frequency probability of p, this displays the Punnet Square Hardy Weinberg frequencies

Hypergeometric Distribution

Free Hypergeometric Distribution Calculator - Calculates the probability of drawing x objects out of a subgroup of k with n possibilities in a total group of N using the hypergeometric distribution.

If 3 coins are flipped simultaneously, the probability of having three tails is

If 3 coins are flipped simultaneously, the probability of having three tails is...
The probability of flipping a head is 1/2. Since each coin flip is independent, we multiply the probabilities together of the three coin flips:
P(HHH) = 1/2 * 1/2 * 1/2
P(HHH) = [B]1/8[/B]

If 4 people have the same 7 shirts, what is the chance that they will wear the same shirt on one day

If 4 people have the same 7 shirts, what is the chance that they will wear the same shirt on one day?
[LIST=1]
[*]For each person, the probability they all wear the first shirt is 1/4 * 1/4 * 1/4 * 1/4 = 1/256
[*]For each person, the probability they all wear the second shirt is 1/4 * 1/4 * 1/4 * 1/4 = 1/256
[*]For each person, the probability they all wear the third shirt is 1/4 * 1/4 * 1/4 * 1/4 = 1/256
[*]For each person, the probability they all wear the fourth shirt is 1/4 * 1/4 * 1/4 * 1/4 = 1/256
[*]For each person, the probability they all wear the fifth shirt is 1/4 * 1/4 * 1/4 * 1/4 = 1/256
[*]For each person, the probability they all wear the sixth shirt is 1/4 * 1/4 * 1/4 * 1/4 = 1/256
[*]For each person, the probability they all wear the seventh shirt is 1/4 * 1/4 * 1/4 * 1/4 = 1/256
[/LIST]
Now, we add up all those probabilities to get our answer, since any of the 7 scenarios above meets the criteria:
(1 + 1 + 1 + 1 + 1 + 1 + 1)/256
[B]7/256[/B]

If A and B are independent events with P(A) = 0.2 and P(B) = 0.6, then P(A U B)=

If A and B are independent events with P(A) = 0.2 and P(B) = 0.6, then P(A U B)=?
We know the following formula for the probability of 2 events:
P(A U B) = P(A) + P(B) - P(A intersection B)
We're told A and B are independent, which makes P(A intersection B) = 0. So we're left with:
P(A U B) = P(A) + P(B) - P(A intersection B)
P(A U B) = 0.2 + 0.6 - 0
P(A U B) = [B]0.8[/B]

If a die is rolled, what is the probability that the number rolled will not be a "5"?

If a die is rolled, what is the probability that the number rolled will not be a "5"?
Possible rolls:
{1, 2, 3, 4, 5, 6}
Probability of not a 5 means:
{1, 2, 3, 4, 6}
P(Not 6) = 1 - P(6)
P(Not 6) = 1 - 1/6
P(Not 6) = [B]5/6[/B]

if flip a coin 4 times, what is the probability of getting all 4 tails

if flip a coin 4 times, what is the probability of getting all 4 tails.
P(Tails) = 1/2
Each flip is independent, so we have:
[URL='https://www.mathcelebrity.com/cointoss.php?hts=TTTT&hct=+2&tct=+1&fct=+5>=no+more+than&nmnl=+2&htpick=tails&calc=1&montect=500&pl=Calculate+Probability']P(TTTT)[/URL] = [B]1/16[/B]

If power is big, you can assume:

If power is big, you can assume:
a. The difference between the means is more likely to be detected
b. The significance level set by the researcher must be high
c. We increase the probability of type I error
d. Your study result will be more likely to be inconclusive
[B]b. The significance level set by the researcher must be high[/B]

If the probability of an event occurring is 7%, what is the probability of an event not occurring?

If the probability of an event occurring is 7%, what is the probability of an event not occurring?
The probability of all event is 1, or 100%.
If we treat the success of an event as p, then q is 1 - p.
Using percentages, we have:
q = 100% - p
Given p = 7%, we have:
q = 100% - 7%
q = [B]93%[/B]

If the probability of getting struck by lighting each year is 1 in 1,000,000, what is the probabilit

If the probability of getting struck by lighting each year is 1 in 1,000,000, what is the probability that you will not be struck by lightning in one year?
Our sample space is either getting struck by lightning or NOT getting struck by lightning. So we have:
P(Not getting struck by lightning) = 1 - P(Getting struck by lightning)
P(Not getting struck by lightning) = 1 - 1/1,000,000
P(Not getting struck by lightning) = [B]999,999/1,000,000[/B]

If the probability of rain is 15%, what is the probability that it won't rain?

If the probability of rain is 15%, what is the probability that it won't rain?
If we assign the probability of raining as event A, then A' (A complement) is the probability it won't rain. Since it either rains or doesn't rain are the only two events.
There exists an axiom in statistics that states:
P(A) + P(A') = 1
Rearranging this, we get:
P(A') = 1 - P(A)
If we assign the probability of raining as event A which is 0.15, we get:
P(A') = 1 - 0.15
P(A') = [B]0.85[/B]

If the probability of winning is X, what is the probability of losing? (Assume there are no ties.)

If the probability of winning is X, what is the probability of losing? (Assume there are no ties.)
This means you can either win or lose. Since all probabilities in the sample space must add up to 1, then we have:
P(Winning) + P(Losing) = 1
P(Losing) = 1 - P(Winning)
Since P(Winning) = X, we have:
P(Losing) = [B]1 - X[/B]

If the probability that you will correctly reject a false null hypothesis is 0.80 at 0.05 significan

If the probability that you will correctly reject a false null hypothesis is 0.80 at 0.05 significance level. Therefore, α is__ and β is__.
[LIST]
[*]α represents the significance level of 0.05
[*]We want the Power of a Test which is 1 - β = 0.8 so β = 0.20
[/LIST]
Our answer is: [B]0.05, 0.20 [/B]

If there are 52 cards in a pack, what is the probability of picking 2 kings in a row when the first

If there are 52 cards in a pack, what is the probability of picking 2 kings in a row when the first card picked is not put back?
4 kings in a deck, and 52 cards in a pack.
First draw, the probability of drawing a king is 4/52.
Second draw, we have 51 cards left since we do not put the first card back, and only 3 Kings left. So the second draw probability for a King is 3/51.
Since each draw is independent, we multiply the first and second draws:
4/52 * 3/51 = [B]12/2652 = 0.0045[/B]

If two coins are flipped, what is the probability that there will not be two heads?

If two coins are flipped, what is the probability that there will not be two heads?
There's only one way to flip 2 coins and get 2 heads:
P(HH) = 1/2 * 1/2 = 1/4
Which means the probability of NOT getting 2 heads is:
1 - 1/4 = [B]3/4
[MEDIA=youtube]vNbA7vE361M[/MEDIA][/B]

If two standard dice are rolled, what is the probability that the sum is 3?

If two standard dice are rolled, what is the probability that the sum is 3?
[URL='https://www.mathcelebrity.com/2dice.php?gl=1&pl=3&opdice=1&rolist=+&dby=&ndby=&montect=+']Using our 2 dice calculator[/URL], we get:
[B]1/18[/B]

if you have a bag with 7 red balls in it and 3 yellow balls in it, whats the probability of pulling

if you have a bag with 7 red balls in it and 3 yellow balls in it, whats the probability of pulling out a yellow ball
P(Yellow) = 3/(3 + 7)
P(Yellow) = [B]3/10 or 0.3[/B].

If you throw a die for two times, what is the probability that you will get a one on the first throw

If you throw a die for two times, what is the probability that you will get a one on the first throw or a one on the second throw (or both)?
[LIST]
[*]P(1) on first roll and P(anything on second roll) = 1/6 * 1 = 1/6
[*]P(anything on first roll) and P(1) on second roll = 1 * 1/6 = 1/6
[*]Add those together: 1/6 + 1/6 = 2/6 = [B]1/3[/B]
[/LIST]

If you toss a fair coin 6 times, what is the probability of getting all tails?

If you toss a fair coin 6 times, what is the probability of getting all tails?
We [URL='https://www.mathcelebrity.com/cointoss.php?hts=TTTTTT&hct=+2&tct=+1&fct=+5>=no+more+than&nmnl=+2&htpick=tails&calc=1&montect=500&pl=Calculate+Probability']type in our search engine [I]TTTTTT [/I]and we get[/URL]:
P(TTTTTT) = [B]1/64 or 0.015625[/B]

Imagine a researcher posed a null hypothesis that in a certain community, the average energy expendi

Imagine a researcher posed a null hypothesis that in a certain community, the average energy expenditure should be 2,100 calories per day. He randomly sampled 100 people in that community. After he computed the t value by calculating a two-tailed t-statistic, he found that the probability value was 0.10. Thus, he concluded:
a. The average energy expenditure was bigger than 2,100 calories per day
b. The average energy expenditure was smaller than 2,100 calories per day
c. He could not reject the null hypothesis that the average energy expenditure was 2,100 calories per day
d. The average energy expenditure was either more than 2,100 calories per day or less than 2,100 calories per day
[B]c. He could not reject the null hypothesis that the average energy expenditure was 2,100 calories per day[/B]
[I]p-value is higher than 0.05[/I]

In a certain Algebra 2 class of 26 students, 18 of them play basketball and 7 of them play baseball.

In a certain Algebra 2 class of 26 students, 18 of them play basketball and 7 of them play baseball. There are 5 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?
Students play either basketball only, baseball only, both sports, or no sports. Let the students who play both sports be b. We have:
b + 18 + 7 - 5 = 26 <-- [I]We subtract 5 because we don't want to double count the students who played a sport who were counted already
[/I]
We [URL='https://www.mathcelebrity.com/1unk.php?num=b%2B18%2B7-5%3D26&pl=Solve']type this equation into our search engine[/URL] and get:
b = [B]6[/B]

In a certain lot, there are 16 white, 7 red, 8 blue, and 9 black cars. You randomly pick a set of ke

In a certain lot, there are 16 white, 7 red, 8 blue, and 9 black cars. You randomly pick a set of keys to one of the cars. What is the probability of choosing a set of keys to a blue car?
[U]Our total cars are:[/U]
Total Cars = White Cars + Red Cars + Blue Cars = Black Cars
Total Cars = 16 + 7 + 8 + 9
Total Cars = 40
P(Blue) = Blue Cars / Total Cars
P(Blue) = 8/40
Using our [URL='https://www.mathcelebrity.com/fraction.php?frac1=8%2F40&frac2=3%2F8&pl=Simplify']fraction simplify calculator[/URL], we get:
P(Blue) = [B]1/5[/B]

In a factory that manufactures tires, a machine responsible for molding the tire has a failure rate

In a factory that manufactures tires, a machine responsible for molding the tire has a failure rate of 0.2%. If 1,000 tires are produced in a day, of which 6 are faulty, what is the difference between the experimental probability and the theoretical probability?
Theoretical probability = Failure Rate * Tires
Theoretical probability = 0.002 * 1000
Theoretical probability = 2
The experimental probability was given as 6, so the difference is:
6 - 2 = [B]4[/B]

In a family of 4 children, what is the probability that all four will be girls?

In a family of 4 children, what is the probability that all four will be girls?
P(G) = 1/2 or 0.5
Since each child is independent, we have:
1/2 * 1/2 * 1/2 * 1/2 or (1/2)^4
[B]1/16 or in decimal form, 0.0625[/B]

In a paper bag, 7 of the 15 marbles are yellow. In a cloth bag, 2 of the 15 marbles are yellow. If

In a paper bag, 7 of the 15 marbles are yellow. In a cloth bag, 2 of the 15 marbles are yellow. If Tim randomly draws one marble from each bag, what is the probability that they are both yellow?
Bag 1 probability of drawing yellow is 7/15
Bag 2 probability of drawing yellow is 2/15
Since each event is independent, we multiply each draw to get our final probability:
P(yellow Bag 1)(yellow Bag 2) = P(Yellow Bag 1) * P(Yellow Bag 2)
P(yellow Bag 1)(yellow Bag 2) = 7/15 * 2/15
P(yellow Bag 1)(yellow Bag 2) = [B]14/225[/B]
[URL='https://www.mathcelebrity.com/fraction.php?frac1=14%2F225&frac2=3%2F8&pl=Simplify']Since we cannot simplify this fraction anymore[/URL], our answer is [B]14/225[/B]

It is estimated that weekly demand for gasoline at new station is normally distributed, with an aver

It is estimated that weekly demand for gasoline at new station is normally distributed, with an average of 1,000 and standard deviation of 50 gallons. The station will be supplied with gasoline once a week. What must the capacity of its tank be if the probability that its supply will be exhausted in a week is to be no more than 0.01?
0.01 is the 99th percentile
Using our [URL='http://www.mathcelebrity.com/percentile_normal.php?mean=+1000&stdev=50&p=99&pl=Calculate+Percentile']percentile calculator[/URL], we get [B]x = 1116.3[/B]

Jennie and Alex both wanted to get a free ticket for a College Music concert. However, the concert s

Jennie and Alex both wanted to get a free ticket for a College Music concert. However, the concert staff told them the tickets were limited. Twenty people wanted to attend the concert but only 10 free tickets were left. So the concert center staff decided to use a lottery to decide who would receive the free tickets. What's the probability of Jennie and Alex both getting free tickets?
1/2 * 1/2 = 1/4 = [B]0.25[/B]

Jennifer is playing cards with her bestie when she draws a card from a pack of 25 cards numbered fro

Jennifer is playing cards with her bestie when she draws a card from a pack of 25 cards numbered from 1 to 25. What is the probability of drawing a number that is square?
The squares from 1 - 25 less than or equal to 25 are as follows:
[LIST=1]
[*]1^2 = 1
[*]2^2 = 4
[*]3^2 = 9
[*]4^2 = 16
[*]5^2 = 25
[/LIST]
So the following 5 cards are squares:
{1, 4, 9, 16, 25}
Therefore, our probability of drawing a square is:
P(square) = Number of Squares / Number of Cards
P(square) = 5/25
This fraction can be simplified. So [URL='https://www.mathcelebrity.com/fraction.php?frac1=5%2F25&frac2=3%2F8&pl=Simplify']we type in 5/25 into our search engine, choose simplify[/URL], and we get:
P(square) = [B]1/5[/B]

Jerry rolls a dice 300 times what is the estimated numbers the dice rolls on 6

Jerry rolls a dice 300 times what is the estimated numbers the dice rolls on 6
Expected Value = Rolls * Probability
Since a 6 has a probability of 1/6, we have:
Expected Value = 300 * 1/6
Expected Value = [B]50[/B]

Larry is rolling two dice. His dad told him that he can skip doing the dishes that night unless he r

Larry is rolling two dice. His dad told him that he can skip doing the dishes that night unless he rolls double sixes. What is the probability that Larry will be able to skip doing the dishes?
P(6, 6) = 1/6 * 1/6 = 1/36
P(Not 6,6) = 1 - 1/36 = [B]35/36[/B]

Let A and B be independent events with P(A) = 0.52 and P(B) = 0.62. a. Calculate P(A ∩ B).

Let A and B be independent events with P(A) = 0.52 and P(B) = 0.62. a. Calculate P(A ∩ B).
With independent events, the intersection probability is found by:
P(A ∩ B) = P(A) * P(B)
P(A ∩ B) = 0.52 * 0.62
P(A ∩ B) = [B]0.3224[/B]

Lotto Drawing Probability

Free Lotto Drawing Probability Calculator - Given a lotto drawing with a Pick(x) out of (y) total choices, this calculates the probability of winning that lottery picking all (x) correct numbers.

mark has a drawer full of 12 yellow baseball caps and 18 white baseball caps. What is the probabilit

mark has a drawer full of 12 yellow baseball caps and 18 white baseball caps. What is the probability of the next cap he chooses at random will be yellow?
P(yellow) = yellow caps / Total caps
P(yellow) = 12/(12 + 18)
P(yellow) = 12/30
[URL='https://www.mathcelebrity.com/fraction.php?frac1=12.%2F30&frac2=3%2F8&pl=Simplify']Simplifying this fraction,[/URL] we get:
P(yellow) = [B]2/5[/B]

Mental Models of Math Book

Calculation Domination: How Anybody Can Explode Their Math Scores Using the Mental Magic of Elon Musk and Warren Buffett.
[MEDIA=youtube]RclG-k6itpk[/MEDIA]
This audiobook shows you various mental models of the top 5% of math students. Mental models come from the following disciplines: Math Anxiety Science Problem Solving Probability Decision Making Metaphysics Persuasion Math Testing

Mimi just started her tennis class three weeks ago. On average, she is able to return 20% of her opp

Mimi just started her tennis class three weeks ago. On average, she is able to return 20% of her opponent's serves. Assume her opponent serves 8 times. Show all work. Let X be the number of returns that Mimi gets. As we know, the distribution of X is a binomial probability distribution.
a) What is the number of trials (n), probability of successes (p) and probability of failures (q), respectively?
b) Find the probability that that she returns at least 1 of the 8 serves from her opponent.
(c) How many serves can she expect to return?
a) [B]n = 8
p = 0.2[/B]
q = 1 - p
q = 1 - 0.2
[B]q = 0.8
[/B]
b) [B]0.4967[/B] on our [URL='http://www.mathcelebrity.com/binomial.php?n=+8&p=0.2&k=1&t=+5&pl=P%28X+>+k%29']binomial calculator[/URL]
c) np = 8(0.2) = 1.6 ~ [B]2[/B] using the link above

Multinomial Distribution

Free Multinomial Distribution Calculator - Given a set of x_{i} counts and a respective set of probabilities θ_{i}, this calculates the probability of those events occurring.

Negative Binomial Distribution

Free Negative Binomial Distribution Calculator - Calculates the probability of the k^{th} success on the x^{th} try for a negative binomial distribution also known as the Pascal distribution.? ? It calculates the probability of exactly k successes, no more than k successes, and greater than k successes as well as the mean, variance, and standard deviation.

Normal Distribution

Free Normal Distribution Calculator - Calculates the probability that a random variable is less than or greater than a value or between 2 values using the Normal Distribution z-score (z value) method (Central Limit Theorem).

Also calculates the Range of values for the 68-95-99.7 rule, or three-sigma rule, or empirical rule. Calculates z score probability

Also calculates the Range of values for the 68-95-99.7 rule, or three-sigma rule, or empirical rule. Calculates z score probability

Odds Probability

Free Odds Probability Calculator - Given an odds prediction m:n of an event success, this calculates the probability that the event will occur or *not* occur

Of all smokers in particular district, 40% prefer brand A and 60% prefer brand B. Of those who prefe

Of all smokers in particular district, 40% prefer brand A and 60% prefer brand B. Of those who prefer brand A, 30% are female, and of those who prefer brand B, 40% are female.
Q: What is the probability that a randomly selected smoker prefers brand A, given that the person selected is a female?
P(F) = P(F|A)*P(A) + P(F|B)*P(B)
P(F) = 0.3*0.4 + 0.4*0.6 = 0.36
So, 36% of all the smokers are female.
You are looking for P(A|F)
P(A|F) = P(A and F)/P(F)
P(A|F) = (P(F|A)*P(A))/P(F)
P(A|F) = (0.3 * 0.4)/0.36
P(A|F) = [B]0.33 or 33%[/B]

Of the 20 boats at the Mariana, 10 were from Massachusetts. What is the probability that a randomly

Of the 20 boats at the Mariana, 10 were from Massachusetts. What is the probability that a randomly selected boat will be from Massachusetts?
P(Boat from Massachusetts) = Number of Massachusetts boats / Total Boats at the Mariana
P(Boat from Massachusetts) = 10/20
[URL='https://www.mathcelebrity.com/fraction.php?frac1=10%2F20&frac2=3%2F8&pl=Simplify']Simplifying this fraction, we get[/URL]:
P(Boat from Massachusetts) = [B]1/2[/B]

On your first draw, what is the probability of drawing a red card, without looking, from a shuffled

On your first draw, what is the probability of drawing a red card, without looking, from a shuffled deck containing 6 red cards, 6 blue cards, and 8 black cards?
P(Red) = Total Red / Total Cards
P(Red) = 6 red/(6 red + 6 blue + 8 black)
P(Red) = 6/20
This fraction can be simplified.
The [URL='https://www.mathcelebrity.com/gcflcm.php?num1=6&num2=20&num3=&pl=GCF+and+LCM']greatest common factor of 6 and 20[/URL] is 2.
So we divide top and bottom of our probability by 2:
P(Red) = 6/2 / 20 / 2
P(Red) = [B]3/10[/B]

Please help!!

(1), how can probability be greater than 1?

Poisson Distribution

Free Poisson Distribution Calculator - Calculates the probability of 3 separate events that follow a poisson distribution.

It calculates the probability of exactly k successes P(x = k)

No more than k successes P (x <= k)

Greater than k successes P(x >= k)

Each scenario also calculates the mean, variance, standard deviation, skewness, and kurtosis.

Calculates moment number t using the moment generating function

It calculates the probability of exactly k successes P(x = k)

No more than k successes P (x <= k)

Greater than k successes P(x >= k)

Each scenario also calculates the mean, variance, standard deviation, skewness, and kurtosis.

Calculates moment number t using the moment generating function

population MEAN OF ENVIRONMENTAL SPECIALIST SALARY IS $62000.A RANDOM SAMPLE OF 45 SPECIALIST IS DRA

population MEAN OF ENVIRONMENTAL SPECIALIST SALARY IS $62000.A RANDOM SAMPLE OF 45 SPECIALIST IS DRAWN FROM THE POPULATION. WHAT IS THE LIKELIHOOD THAT THE MEAN SALARY SAMPLE IS $59000. ASSUME SIGMA IS $6000.
Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=59000&mean=62000&stdev=6000&n=45&pl=P%28X+%3C+Z%29']Z-Score calculator[/URL], we get the probability as [B]0.0004[/B].

Probability

Free Probability Calculator - This lesson walks you through the basics of probability like the probability definition, events, outcomes, experiments, and probability postulates

Probability (A U B U C)

Free Probability (A U B U C) Calculator - Calculates the probability of a union of a three event sample space, A, B, and C, as well as P(A), P(B), P(C), P(A ∩ B), P(A ∩ C), P(B ∩ C), P(A ∩ B ∩ C).

Probability (A U B)

Free Probability (A U B) Calculator - Given a 2 event sample space A and B, this calculates the probability of the following events:

P(A U B)

P(A)

P(B)

P(A ∩ B)

P(A U B)

P(A)

P(B)

P(A ∩ B)

Probability of getting 4 or 6 when rolling a dice

Probability of getting 4 or 6 when rolling a dice
P(4 or 6) = P(4) + P(6)
P(4 or 6) = 1/6 + 1/6
P(4 or 6) = 2/6
We can simplify this. We [URL='https://www.mathcelebrity.com/fraction.php?frac1=2%2F6&frac2=3%2F8&pl=Simplify']type this fraction into our search engine, choose simplify[/URL], and we get:
P(4 or 6) = [B]1/3[/B]

Random Sampling from the Normal Distribution

Free Random Sampling from the Normal Distribution Calculator - This performs hypothesis testing on a sample mean with critical value on a sample mean or calculates a probability that Z <= z or Z >= z using a random sample from a normal distribution.

Refer to a bag containing 13 red balls numbered 1-13 and 5 green balls numbered 14-18. You choose a

Refer to a bag containing 13 red balls numbered 1-13 and 5 green balls numbered 14-18. You choose a ball at random.
a. What is the probability that you choose a red or even numbered ball?
b. What is the probability you choose a green ball or a ball numbered less than 5?
a. The phrase [I]or[/I] in probability means add. But we need to subtract even reds so we don't double count:
We have 18 total balls, so this is our denonminator for our fractions.
Red and Even balls are {2, 4, 6, 8, 10, 12}
Our probability is:
P(Red or Even) = P(Red) + P(Even) - P(Red and Even)
P(Red or Even) = 13/18 + 9/18 - 6/18
P(Red or Even) = 16/18
Using our [URL='https://www.mathcelebrity.com/fraction.php?frac1=16%2F18&frac2=3%2F8&pl=Simplify']Fraction Simplify Calculator[/URL], we have:
P(Red or Even) = [B]16/18[/B]
[B][/B]
b. The phrase [I]or[/I] in probability means add. But we need to subtract greens less than 5 so we don't double count:
We have 18 total balls, so this is our denonminator for our fractions.
Green and less than 5 does not exist, so we have no intersection
Our probability is:
P(Green or Less Than 5) = P(Green) + P(Less Than 5) - P(Green And Less Than 5)
P(Green or Less Than 5) = 5/18 + 4/18 - 0
P(Green or Less Than 5) = 9/18
Using our [URL='https://www.mathcelebrity.com/fraction.php?frac1=9%2F18&frac2=3%2F8&pl=Simplify']Fraction Simplify Calculator[/URL], we have:
P(Red or Even) = [B]1/2[/B]

Roulette

Free Roulette Calculator - Calculates the probability for different bets on a roulette wheel including expected return on a monetary bet.

Roulette Cumulative Betting

Free Roulette Cumulative Betting Calculator - This calculator displays the probability and return grid for a roulette scenario where you play x games, betting y per number playing z numbers per game.

Rule of Succession

Free Rule of Succession Calculator - Given s successes in n independent trials, this calculates the probability that the next repetition is a success

Sample Space Probability

Free Sample Space Probability Calculator - Given a sample space S and an Event Set E, this calculates the probability of the event set occuring.

Sara has a box of candies. In the box there are 8 pink candies, 7 purple candies and 5 blue candies.

Sara has a box of candies. In the box there are 8 pink candies, 7 purple candies and 5 blue candies. She takes one candy and records its color. She then puts it back in the box and draws another candy. What is the probability of taking out a pink candy followed by a blue candy?
[B][U]Calculate the total number of candies:[/U][/B]
Total candies = Pink + Purple + Blue
Total candies = 8 + 7 + 5
Total candies = 20
[B][U]Calculate the probability of drawing one pink candy:[/U][/B]
P(Pink) = 8/20
Using our [URL='https://www.mathcelebrity.com/fraction.php?frac1=8%2F20&frac2=3%2F8&pl=Simplify']fraction reduction calculator[/URL], we get:
P(Pink) = 2/5
[B][U]Calculate the probability of drawing one blue candy:[/U][/B]
P(Blue) = 5/20 <-- [I]20 options since Sara replaced her first draw[/I]
Using our [URL='https://www.mathcelebrity.com/fraction.php?frac1=5%2F20&frac2=3%2F8&pl=Simplify']fraction reduction calculator[/URL], we get:
P(Blue) = 1/4
The problem asks for the probability of a Pink followed by a Blue. Since each event is independent, we multiply:
P(Pink, Blue) = P(Pink) * P(Blue)
P(Pink, Blue) = 2/5 * 1/4
P(Pink, Blue) = 2/20
Using our [URL='https://www.mathcelebrity.com/fraction.php?frac1=2%2F20&frac2=3%2F8&pl=Simplify']fraction reduction calculator[/URL], we get:
P(Pink, Blue) = [B]1/10 or 10%[/B]

Sarah rolls 2 fair dice and adds the results from each. Work out the probability of getting a total

Sarah rolls 2 fair dice and adds the results from each. Work out the probability of getting a total that is a factor of 6.
Factors of 6 are {6, 12}
[URL='https://www.mathcelebrity.com/2dice.php?gl=1&pl=6&opdice=1&rolist=+&dby=&ndby=&montect=+']P(Roll a 6)[/URL] = 5/36
[URL='https://www.mathcelebrity.com/2dice.php?gl=1&pl=12&opdice=1&rolist=+&dby=&ndby=&montect=+']P(Roll a 12)[/URL] = 1/36
P(Roll a 6 or Roll a 12) = P(Roll a 6) + P(Roll a 12)
P(Roll a 6 or Roll a 12) = 5/36 + 1/36
P(Roll a 6 or Roll a 12) = 6/36
Using our [URL='https://www.mathcelebrity.com/fraction.php?frac1=6%2F36&frac2=3%2F8&pl=Simplify']fraction simplifier[/URL], we see that:
P(Roll a 6 or Roll a 12) = [B]1/6[/B]

Standard Normal Distribution

Free Standard Normal Distribution Calculator - Givena normal distribution z-score critical value, this will generate the probability. Uses the NORMSDIST Excel function.

Success in a binomial event is .15 what is the probability of failure?

Success in a binomial event is .15 what is the probability of failure?
Success is represented as p. p = 0.15.
The probability of failure q, is written as q = 1 - p
q = 1 - 0.15
[B]q = 0.85[/B]

Suppose that 18% of people own dogs. If you pick two people at random, what is the probability that

Suppose that 18% of people own dogs. If you pick two people at random, what is the probability that they both own a dog?
Since each person is independent of the others, we have:
P(Person 1 has a dog and person 2 have a dog) = P(person 1 has a dog) * P(person 2 has a dog)
P(Person 1 has a dog and person 2 have a dog) = 0.18 * 0.18
P(Person 1 has a dog and person 2 have a dog) = [B]0.0324 or 3.24%[/B]

Suppose that previously collected traffic data indicate that, during the afternoon rush hour, an ave

Suppose that previously collected traffic data indicate that, during the afternoon rush hour, an average of 4 cars arrive at a toll bridge each second. If it is assumed that cars arrive randomly, and can thus be modeled with Poisson distribution, what is the probability that in the next second, [U][B]NO[/B][/U] cars will arrive?
Use the [I]Poisson Distribution[/I] with λ = 4 and x = 0
Using the [URL='http://www.mathcelebrity.com/poisson.php?n=+10&p=+0.4&k=+0&t=+3&pl=P%28X+=+k%29']Poisson Distribution calculator[/URL], we get P(0; 4) = [B]0.0183[/B]

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed wit

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls.
a. If X = average distance in feet for 49 fly balls, then X ~ _______(_______,_______)

b. What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for X. Shade the region corresponding to the probability. Find the probability.

c. Find the 80^{th} percentile of the distribution of the average of 49 fly balls
a. N(250, 50/sqrt(49)) = [B]0.42074[/B]
b. Calculate Z-score and probability = 0.08 shown [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+240&mean=+250&stdev=+7.14&n=+1&pl=P%28X+%3C+Z%29']here[/URL]
c. Inverse of normal distribution(0.8) = 0.8416. Use NORMSINV(0.8) [URL='http://www.mathcelebrity.com/zcritical.php?a=0.8&pl=Calculate+Critical+Z+Value']calculator[/URL]
Using the Z-score formula, we have
0.8416 = (x - 250)/50
x = [B]292.08[/B]

b. What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for X. Shade the region corresponding to the probability. Find the probability.

c. Find the 80

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed wit

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and
a standard deviation of 50 feet.
a. If X = distance in feet for a fly ball, then X ~
b. If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 220 feet? Sketch the graph. Scale the horizontal axis X. Shade the region corresponding to the probability. Find the probability.
c. Find the 80th percentile of the distribution of fly balls. Sketch the graph, and write the probability statement.
a. [B]N(250, 50/sqrt(1))[/B]
b. Calculate [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+220&mean=250&stdev=50&n=+1&pl=P%28X+%3C+Z%29']z-score[/URL]
Z = -0.6 and P(Z < -0.6) = [B]0.274253[/B]
c. Inverse of normal distribution(0.8) = 0.8416 using NORMSINV(0.8) [URL='http://www.mathcelebrity.com/zcritical.php?a=0.8&pl=Calculate+Critical+Z+Value']calculator[/URL]
Z-score formula: 0.8416 = (x - 250)/50

x = [B]292.08[/B]

x = [B]292.08[/B]

Suppose that the manager of the Commerce Bank at Glassboro determines that 40% of all depositors hav

Suppose that the manager of the Commerce Bank at Glassboro determines that 40% of all depositors have a multiple accounts at the bank. If you, as a branch manager, select a random sample of 200 depositors, what is the probability that the sample proportion of depositors with multiple accounts is between 35% and 50%?
[URL='http://www.mathcelebrity.com/proportion_hypothesis.php?x=50&n=+100&ptype==&p=+0.4&alpha=+0.05&pl=Proportion+Hypothesis+Testing']50% proportion probability[/URL]: z = 2.04124145232
[URL='http://www.mathcelebrity.com/proportion_hypothesis.php?x=+35&n=+100&ptype==&p=+0.4&alpha=+0.05&pl=Proportion+Hypothesis+Testing']35% proportion probability[/URL]: z = -1.02062072616
Now use the [URL='http://www.mathcelebrity.com/zscore.php?z=p%28-1.02062072616

Suppose two number cubes are rolled. What is the probability of rolling a sum greater than 8?

List out the sums greater than 8:
(4, 5)
(4, 6)
(5, 5)
(5, 6)
(6, 6)
(5, 4)
(6, 4)
(6, 5)
Since there are 6 * 6 = 36 total outcomes, we have the probability of the sum greater than 8 as:
8/36 = 2/9

Survival Rates

Free Survival Rates Calculator - Given a set of times and survival population counts, the calculator will determine the following:

Survival Population l_{x}

Mortality Population d_{x}

Survival Probability p_{x}

Mortality Probability q_{x}

In addition, the calculator will determine the probability of survival from t_{x} to t_{x + n}

Survival Population l

Mortality Population d

Survival Probability p

Mortality Probability q

In addition, the calculator will determine the probability of survival from t

Tabular Display

Free Tabular Display Calculator - Enter a set of x and p(x) in a tabular probability distribution format and this will evaluate if it is valid or not.

Taylor is playing a game using a die and a spinner. The spinner is divided into 4 equal parts with c

Taylor is playing a game using a die and a spinner. The spinner is divided into 4 equal parts with colors green, red, yellow, and purple. Taylor rolls the die and spins the spinner. What is the probability the die shows a 2 and the spinner lands on purple?
Probability of rolling a 2 on the die is 1/6
Probability of getting a purple on the spinner is 1/4
Since each event is independent, our joint probability is:
P(2 on the die and Purple on the spinner) = P(2 on the die) x P(Purple on the Spinner)
P(2 on the die and Purple on the spinner) = 1/6 x 1/4
P(2 on the die and Purple on the spinner) = [B]1/24[/B]

The average precipitation for the first 7 months of the year is 19.32 inches with a standard deviati

The average precipitation for the first 7 months of the year is 19.32 inches with a standard deviation of 2.4 inches. Assume that the average precipitation is normally distributed.
a. What is the probability that a randomly selected year will have precipitation greater than 18 inches for the first 7 months?
b. What is the average precipitation of 5 randomly selected years for the first 7 months?
c. What is the probability of 5 randomly selected years will have an average precipitation greater than 18 inches for the first 7 months?
[URL='https://www.mathcelebrity.com/probnormdist.php?xone=18&mean=19.32&stdev=2.4&n=1&pl=P%28X+%3E+Z%29']For a. we set up our z-score for[/URL]:
P(X>18) = 0.7088
b. We assume the average precipitation of 5 [I]randomly[/I] selected years for the first 7 months is the population mean μ = 19.32
c. [URL='https://www.mathcelebrity.com/probnormdist.php?xone=18&mean=19.32&stdev=2.4&n=5&pl=P%28X+%3E+Z%29']P(X > 18 with n = 5)[/URL] = 0.8907

The chance of a soldier being an enemy spy is .0005. Out of 10,000 soldiers, how many of them are ex

The chance of a soldier being an enemy spy is .0005. Out of 10,000 soldiers, how many of them are expected to be spies?
Expected Spies = Probability of being a spy * Total Soldiers
Expected Spies = 0.0005 * 10000
Expected Spies = [B]5[/B]

The distribution of actual weights of 8 oz chocolate bars produced by a certain machine is normal wi

The distribution of actual weights of 8 oz chocolate bars produced by a certain machine is normal with µ=8.1 ounces and σ=0.1 ounces. A sample of 5 of these chocolate bars is selected. What is the probability that their average weight is less than 8 ounces?
Calculate Z score and probability using [URL='http://www.mathcelebrity.com/probnormdist.php?xone=8&mean=8.1&stdev=0.1&n=5&pl=P%28X+%3C+Z%29']our calculator[/URL]:
Z = -2.236
P(X < -2.236) = [B]0.012545[/B]

The first significant digit in any number must be 1, 2, 3, 4, 5, 6, 7, 8, or 9. It was discovered t

The first significant digit in any number must be 1, 2, 3, 4, 5, 6, 7, 8, or 9. It was discovered that first digits do not occur with equal frequency. Probabilities of occurrence to the first digit in a number are shown in the accompanying table. The probability distribution is now known as Benford's Law. For example, the following distribution represents the first digits in 231 allegedly fraudulent checks written to a bogus company by an employee attempting to embezzle funds from his employer.
Digit, Probability
1, 0.301
2, 0.176
3, 0.125
4, 0.097
5, 0.079
6, 0.067
7, 0.058
8, 0.051
9, 0.046
[B][U]Fradulent Checks[/U][/B]
Digit, Frequency
1, 36
2, 32
3, 45
4, 20
5, 24
6, 36
7, 15
8, 16
9, 7
Complete parts (a) and (b).
(a) Using the level of significance α = 0.05, test whether the first digits in the allegedly fraudulent checks obey Benford's Law. Do the first digits obey the Benford's Law?

Yes or No Based on the results of part (a), could one think that the employe is guilty of embezzlement? Yes or No Show frequency percentages Digit Fraud Probability Benford Probability 1 0.156 0.301 2 0.139 0.176 3 0.195 0.125 4 0.087 0.097 5 0.104 0.079 6 0.156 0.067 7 0.065 0.058 8 0.069 0.051 9 0.03 0.046 Take the difference between the 2 values, divide it by the Benford's Value. Sum up the squares to get the Test Stat of 2.725281277 Critical Value Excel: =CHIINV(0.95,8) = 2.733 Since test stat is less than critical value, we cannot reject, so [B]YES[/B], it does obey Benford's Law and [B]NO[/B], there is not enough evidence to suggest the employee is guilty of embezzlement.

Yes or No Based on the results of part (a), could one think that the employe is guilty of embezzlement? Yes or No Show frequency percentages Digit Fraud Probability Benford Probability 1 0.156 0.301 2 0.139 0.176 3 0.195 0.125 4 0.087 0.097 5 0.104 0.079 6 0.156 0.067 7 0.065 0.058 8 0.069 0.051 9 0.03 0.046 Take the difference between the 2 values, divide it by the Benford's Value. Sum up the squares to get the Test Stat of 2.725281277 Critical Value Excel: =CHIINV(0.95,8) = 2.733 Since test stat is less than critical value, we cannot reject, so [B]YES[/B], it does obey Benford's Law and [B]NO[/B], there is not enough evidence to suggest the employee is guilty of embezzlement.

The hourly wages of employees at Rowan have a mean wage rate of $10 per hour with a standard deviati

The hourly wages of employees at Rowan have a mean wage rate of $10 per hour with a standard deviation of $1.20. What is the probability the mean hourly wage of a random sample of 36 employees will be larger than $10.50? Assume the company has a total of 1,000 employees
Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=10.5&mean=10&stdev=1.2&n=36&pl=P%28X+>+Z%29']normal distribution calculator[/URL], we get P(x > 10.5) = [B]0.00621[/B]

The IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. a) What i

The IQ scores are normally distributed with a mean of 100 and a standard deviation of 15.
a) What is the probability that a randomly person has an IQ between 85 and 115?
b) Find the 90th percentile of the IQ distribution
c) If a random sample of 100 people is selected, what is the standard deviation of the sample mean?
a) [B]68%[/B] from the [URL='http://www.mathcelebrity.com/probnormdist.php?xone=50&mean=100&stdev=15&n=1&pl=Empirical+Rule']empirical rule calculator[/URL]
b) P(z) = 0.90. so z = 1.28152 using Excel NORMSINV(0.9)

(X - 100)/10 = 1.21852 X = [B]113[/B] rounded up c) Sample standard deviation is the population standard deviation divided by the square root of the sample size 15/sqrt(100) = 15/10 =[B] 1.5[/B]

(X - 100)/10 = 1.21852 X = [B]113[/B] rounded up c) Sample standard deviation is the population standard deviation divided by the square root of the sample size 15/sqrt(100) = 15/10 =[B] 1.5[/B]

The letters that form the word ALGEBRA are placed in a bowl. What is the probability of choosing a l

The letters that form the word ALGEBRA are placed in a bowl. What is the probability of choosing a letter that is not “A”?
ALGEBRA has 7 letters
Of the 7 letters, we have 2 A's.
So we have 7 - 2 = 5 letters which are not A
P(Not A) = Letters not A / Total letters
P(Not A) = [B]5/7[/B]

The Oakdale High School Speech and Debate Club hosted its annual car wash fundraiser. Each club memb

The Oakdale High School Speech and Debate Club hosted its annual car wash fundraiser. Each club member brought a bottle of car wash soap, so there were 8 total bottles. 6 of the bottles contained orange soap. If a club member randomly selects 5 bottles to pour into the first soap bucket, what is the probability that all of them contain orange soap?
This is assumed to be draw without replacement, so we have:
[LIST=1]
[*]Draw 1: 6/8
[*]Draw 2: 5/7
[*]Draw 3: 4/6
[*]Draw 4: 3/5
[*]Draw 5: 2/4
[/LIST]
Since they are independent events, we multiply:
6/8 * 5/7 * 4/6 * 3/5 * 2/4
(6 * 5 * 4 * 3 * 2)/(8 * 7 * 6 * 5 * 4)
720/6720
[B]0.1071[/B]

THE PLAYER CHOSE 20 OUT OF 70 NUMBERS IN A GAME OF CHANCE. ...WHEN THE SHOW BEGIN,THE BANKER WILL

THE PLAYER CHOSE 20 OUT OF 70 NUMBERS IN A GAME OF CHANCE. ...WHEN THE SHOW BEGIN,THE BANKER WILL THEN RAFFLE OR DO A DRAW WHERE IN THE BANKER PICKS AS WELL 20 OUT OF 70 NUMBERS. .....NOW HERES THE TRICK, FOR YOU TO BEAT THE BANKER .YOUR CHOSEN 20 NUMBERS SHOULD NOT MATCH ANY OF THE BANKER 20 0UT OF 70 NUMBERS THAT HAD BEEN DRAWS IN THE GAME OF SHOW. IF THE 20 NUMBERS YOU HAVE ARE TOTALLY DIFFERENT FROM THE BANKERS 20 NUMBERS DRAWN THEN YOU WIN THE PRICE.
Banker Draw Numbers not matching Total numbers Probability Probability Decimal Cumulative Probability
1 50 70 50/70 0.7142857143 0.7142857143
2 49 69 49/69 0.7101449275 0.5072463768
3 48 68 48/68 0.7058823529 0.358056266
4 47 67 47/67 0.7014925373 0.2511737985
5 46 66 46/66 0.696969697 0.1750605262
6 45 65 45/65 0.6923076923 0.1211957489
7 44 64 44/64 0.6875 0.0833220774
8 43 63 43/63 0.6825396825 0.05687062425
9 42 62 42/62 0.6774193548 0.03852526159
10 41 61 41/61 0.6721311475 0.02589402828
11 40 60 40/60 0.6666666667 0.01726268552
12 39 59 39/59 0.6610169492 0.01141092772
13 38 58 38/58 0.6551724138 0.007476125057
14 37 57 37/57 0.649122807 0.004852923282
15 36 56 36/56 0.6428571429 0.003119736396
16 35 55 35/55 0.6363636364 0.001985286797
17 34 54 34/54 0.6296296296 0.001249995391
18 33 53 33/53 0.6226415094 0.000778299017
19 32 52 32/52 0.6153846154 0.0004789532412
20 31 51 31/51 0.6078431373 [B]0.0002911284407 [/B]

The probability of failing to reject a false null hypothesis is ____

The probability of failing to reject a false null hypothesis is ____
a. α
b. 1 - α
c. 1 - β
d. β
[B]d. β[/B]

The singular form of the word "dice" is "die". Tom was throwing a six-sided die. The first time he t

The singular form of the word "dice" is "die". Tom was throwing a six-sided die. The first time he threw, he got a three; the second time he threw, he got a three again. What's the probability of getting a three at the third time?
Since all trials are independent:
1/6 * 1/6 * 1/6 = [B]1/216[/B]

The weight of a 9.5-inch by 6-inch paperback book published by Leaden Publications, Inc., is 16.2 oz

The weight of a 9.5-inch by 6-inch paperback book published by Leaden Publications, Inc., is 16.2 oz. The standard deviation is 2.9 oz. What is the probability that the average weight of a sample of 33 such books is less than 15.89 oz?
Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=15.89&mean=16.2&stdev=2.9&n=33&pl=P%28X+%3C+Z%29']z-score calculator[/URL], we get:
[B]0.271[/B]

There are 10 true or false questions on a test. You do not know the answer to 4 of the questions, so

There are 10 true or false questions on a test. You do not know the answer to 4 of the questions, so you guess. What is the probability that you will get all 4 answers right?
Probability you guess right is 1/2 or 0.5.
Since each event is independent of the other events, we multiply 1/2 4 times:
1/2 * 1/2 * 1/2 * 1/2 = [B]1/16[/B]

There are 100 people in a sport centre. 67 people use the gym. 62 people use the swimming pool. 5

There are 100 people in a sport centre. 67 people use the gym. 62 people use the swimming pool. 56 people use the track. 38 people use the gym and the pool. 31 people use the pool and the track. 33 people use the gym and the track. 16 people use all three facilities. A person is selected at random. What is the probability that this person doesn't use any facility?
WE use the compound probability formula for 3 events:
[LIST=1]
[*]Gym use (G)
[*]Swimming pool use (S)
[*]Track (T)
[/LIST]
P(G U S U T) = P(G) + P(S) + P(T) - P(G Intersection S) - P(G Intersection T) - P(S Intersection T) + P(G Intersection S Intersection T)
[LIST]
[*]Note: U means Union (Or) and Intersection means (And)
[/LIST]
Plugging our numbers in:
P(G U S U T) = 67/100 + 62/100 + 56/100 - 38/100 - 31/100 - 33/100 + 16/100
P(G U S U T) = (67 + 62 + 56 - 38 - 31 - 33 + 16)/100
P(G U S U T) = 99/100 or 0.99
What this says is, the probability that somebody uses at any of the 3 facilities is 99/100.
The problem asks for none of the 3 facilities, or P(G U S U T)'
P(G U S U T)' = 1 - P(G U S U T)
P(G U S U T)' = 1 - 99/100
P(G U S U T)' = 100/100 - 99/100
P(G U S U T)' = [B]1/100 or 0.1[/B]

There are 1000 juniors in a college. Among the 1000 juniors, 200 students are taking STAT200, and 10

There are 1000 juniors in a college. Among the 1000 juniors, 200 students are taking STAT200, and 100 students are taking PSYC300. There are 50 students taking both courses.
a) What is the probability that a randomly selected junior is taking at least one of these two courses?
b) What is the probability that a randomly selected junior is taking PSYC300, given that he/she is taking STAT200?
a) P(A U B) = P(A) + P(B) - P(A ∩ B) = 0.2 + 0.1 - 0.05 = [B]0.25[/B]
b) P(SYC|STAT) = P(STAT ∩ SYC)/P(STAT) = 0.05/0.2 = [B]0.25[/B]

There are 113 identical plastic chips numbered 1 through 113 in a box. What is the probability of re

There are 113 identical plastic chips numbered 1 through 113 in a box. What is the probability of reaching into the box and randomly drawing a chip number that is greater than 44?
We want 45, 46, … 113
The formula to get inclusive number count between and including 2 numbers is:
Total numbers = L - S + 1
Total numbers = 113 - 45 + 1
Total numbers = 69
That is 69 possible numbers. We draw this out of a total of 113
[B]P(Number > 44) = 69/113
[B]P(Number > 44) [/B]= 0.610619
[MEDIA=youtube]BLBVcpdHqXU[/MEDIA][/B]

There are 15 cards, numbered 1 through 15. If you pick a card, what is the probability that you choo

There are 15 cards, numbered 1 through 15. If you pick a card, what is the probability that you choose an odd number or a two?
We want the P(odd) or P(2).
P(odd) = 1, 3, 5, 7, 9, 11, 13, 15 = 8/15
P(2) = 1/15
Add them both:
8/15 + 1/15 = 9/15
Simplified, we get [B]3/5[/B].

There are 5 orange books, 12 black books, and 8 tan books on Mr. Johnsons bookshelf. Calculate the p

There are 5 orange books, 12 black books, and 8 tan books on Mr. Johnsons bookshelf. Calculate the probability of randomly selecting a black book and then a tan book without replacement. Write your answer as a percent.
P(black book first draw)
P(black book first draw) = 12 black / (5 orange + 12 black + 8 tan)
P(black book first draw) = 12 / 25
P(tan book second draw)
P(tan book second draw) = 8 tan / (5 orange + 11 black + 8 tan) <-- 11 black because we already drew one black
P(tan book second draw) = 8 / 24
Using our fraction reduction calculator, this simplifies to 1/3
Since each draw is independent, we multiply both probabilities:
P(black book first draw, tan book second draw) = 12/25 * 1/3
P(black book first draw, tan book second draw) = 12/75
P(black book first draw, tan book second draw) = [B]16%[/B]

There are 5 red and 4 black balls in a box. If you pick out 2 balls without replacement, what is the

There are 5 red and 4 black balls in a box. If you pick out 2 balls without replacement, what is the probability of getiing at least one red ball?
First list out our sample space. At least one means 1 or 2 red balls, so we have 3 possible draws:
[LIST=1]
[*]Red, Black
[*]Black, Red
[*]Red, Red
[/LIST]
List out the probabilities:
[LIST=1]
[*]Red (5/9) * Black (4/8) = 5/18
[*]Black (4/9) * Red (5/8) = 5/18
[*]Red (5/9) * Red (4/8) = 5/18
[/LIST]
Add these up:
3(5)/18 = [B]5/6[/B]

There is a bag filled with 3 blue, 4 red and 5 green marbles. A marble is taken at random from the

There is a bag filled with 3 blue, 4 red and 5 green marbles. A marble is taken at random from the bag, the colour is noted and then it is not replaced. Another marble is taken at random. What is the probability of getting exactly 1 green?
Calculate Total marbles
Total marbles = Blue + Red + Green
Total marbles = 3 + 4 + 5
Total marbles = 12
Probability of a green = 5/12
Probability of not green = 1 - 5/12 = 7/12
To get exactly one green in two draws, we either get a green, not green, or a not green, green
[U]First Draw Green, Second Draw Not Green[/U]
[LIST]
[*]1st draw: Probability of a green = 5/12
[*]2nd draw: Probability of not green = 7/11 <-- 11 since we did not replace the first marble
[*]To get the probability of the event, since each draw is independent, we multiply both probabilities
[*]Probability of the event is (5/12) * (7/11) = 35/132
[/LIST]
[U]First Draw Not Green, Second Draw Not Green[/U]
[LIST]
[*]1st draw: Probability of not a green = 7/12
[*]2nd draw: Probability of not green = 5/11 <-- 11 since we did not replace the first marble
[*]To get the probability of the event, since each draw is independent, we multiply both probabilities
[*]Probability of the event is (7/12) * (5/11) = 35/132
[/LIST]
To get the probability of exactly one green, we add both of the events:
First Draw Green, Second Draw Not Green + First Draw Not Green, Second Draw Not Green
35/132 + 35/132 = 70/132
[URL='https://www.mathcelebrity.com/fraction.php?frac1=70%2F132&frac2=3%2F8&pl=Simplify']Using our fraction simplify calculator[/URL], we get:
[B]35/66[/B]

There is a bag filled with 4 blue, 3 red and 5 green marbles. A marble is taken at random from the

There is a bag filled with 4 blue, 3 red and 5 green marbles. A marble is taken at random from the bag, the colour is noted and then it is replaced. Another marble is taken at random. What is the probability of getting 2 blues?
We have (4 blue + 3 red + 5 green) = 12 total marbles
With replacement, the probability of getting one blue is 4/12 = 1/3
Since each draw is independent of the last, the probability of Blue, Blue = 1/3 * 1/3 = [B]1/9[/B]

There is a bag filled with 5 blue, 6 red and 2 green marbles. A marble is taken at random from the b

There is a bag filled with 5 blue, 6 red and 2 green marbles. A marble is taken at random from the bag, the colour is noted and then it is replaced. Another marble is taken at random. What is the probability of getting exactly 1 blue?
Find the total number of marbles in the bag:
Total marbles = 5 blue + 6 red + 2 green
Total marbles = 13
The problem asks for exactly one blue in 2 draws [I]with replacement[/I]. Which means you could draw as follows:
Blue, Not Blue
Not Blue, Blue
The probability of drawing a blue is 5/13, since we replace the marbles in the bag each time.
The probability of not drawing a blue is (6 + 2)/13 = 8/13
And since each of the 2 draws are independent of each other, we multiply the probability of each draw:
Blue, Not Blue = 5/13 * 8/13 =40/169
Not Blue, Blue = 8/13 * 5/13 = 40/169
We add both probabilities since they both count under our scenario:
40/169 + 40/169 = 80/169
Checking our [URL='https://www.mathcelebrity.com/fraction.php?frac1=80%2F169&frac2=3%2F8&pl=Simplify']fraction simplification calculator[/URL], we see you cannot simplify this fraction anymore.
So our probability stated in terms of a fraction is 80/169
[URL='https://www.mathcelebrity.com/perc.php?num=80&den=169&pcheck=1&num1=16&pct1=80&pct2=70&den1=80&idpct1=10&hltype=1&idpct2=90&pct=82&decimal=+65.236&astart=12&aend=20&wp1=20&wp2=30&pl=Calculate']Stated in terms of a decimal[/URL], it's 0.4734

There is a stack of 10 cards, each given a different number from 1 to 10. suppose we select a card r

There is a stack of 10 cards, each given a different number from 1 to 10. Suppose we select a card randomly from the stack, replace it, and then randomly select another card. What is the probability that the first card is an odd number and the second card is greater than 7.
First Event: P(1, 3, 5, 7, 9) = 5/10 = 1/2 or 0.5
Second Event: P(8, 9, 10) = 3/10 or 0.3
Probability of both events since each is independent is 1/2 * 3/10 = 3/20 = [B]0.15 or 15%[/B]

three coins are tossed.how many different ways can they fall?

three coins are tossed.how many different ways can they fall?
[URL='https://www.mathcelebrity.com/cointoss.php?hts=+HTHTHH&hct=+2&tct=+1&fct=+5>=no+more+than&nmnl=+2&htpick=heads&tossct=3&montect=3&calc=5&pl=Calculate+Probability']8 outcomes using our coin toss calculator[/URL]

Three ordinary dice are rolled. What is the probability that the results are all less than 5

Three ordinary dice are rolled. What is the probability that the results are all less than 5
Calculate individual die probabilities:
[LIST]
[*]Die 1 P(x < 5) = 4/6 = 2/3
[*]Die 2 P(x < 5) = 4/6 = 2/3
[*]Die 3 P(x < 5) = 4/6 = 2/3
[/LIST]
Since each roll is independent, we have:
P(Die 1 < 5, Die 2 < 5, Die 3 < 5) = 2/3 * 2/3 * 2/3
P(Die 1 < 5, Die 2 < 5, Die 3 < 5) = [B]8/27[/B]

True or False (a) The normal distribution curve is always symmetric to its mean. (b) If the variance

True or False
(a) The normal distribution curve is always symmetric to its mean.
(b) If the variance from a data set is zero, then all the observations in this data set are identical.
(c) P(A AND A^{c})=1, where A^{c} is the complement of A.
(d) In a hypothesis testing, if the p-value is less than the significance level α, we do not have sufficient evidence to reject the null hypothesis.
(e) The volume of milk in a jug of milk is 128 oz. The value 128 is from a discrete data set.
[B](a) True, it's a bell curve symmetric about the mean
(b) True, variance measures how far a set of numbers is spread out. A variance of zero indicates that all the values are identical
(c) True. P(A) is the probability of an event and P(Ac) is the complement of the event, or any event that is not A. So either A happens or it does not. It covers all possible events in a sample space.
(d) False, we have sufficient evidence to reject H0.
(e) False. Volume can be a decimal or fractional. There are multiple values between 127 and 128. So it's continuous.[/B]

Two dice are rolled. Determine the probability of the following. Rolling an even number or a number

Two dice are rolled. Determine the probability of the following. Rolling an even number or a number greater than 6
We want P(X = Even) or P(X>6)
With 2 dice, our die totals are:
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Evens are: 2, 4, 6, 8, 10, 12
> 6 = 7, 8, 9, 10, 11, 12
When the problem states [I]or[/I] it means either of the sets. When we take the union of both sets, we get:
2,4,6,7,8,9,10,11,12
This is 9 possible entries out of 12:
9/12
We can simplify this by dividing top and bottom by 3:
P(X = Even) or P(X>6) = [B]3/4 or 0.75[/B]

two unbiased dice are thrown. find the probability that the total number on the dice is greater than

two unbiased dice are thrown. find the probability that the total number on the dice is greater than 10
[URL='http://www.mathcelebrity.com/2dice.php?gl=2&pl=10&opdice=1&rolist=+2%2C3%2C9%2C10&dby=2%2C3%2C5&ndby=4%2C5&montect=+100']From our 2 dice calculator[/URL]:
We have (5,6),(6,5),(6,6)
P(Sum) > 10 is [B]1/12[/B]

Uniform Distribution

Free Uniform Distribution Calculator - This calculates the following items for a uniform distribution

* Probability Density Function (PDF) ƒ(x)

* Cumulative Distribution Function (CDF) F(x)

* Mean, Variance, and Standard Deviation

Calculates moment number t using the moment generating function

* Probability Density Function (PDF) ƒ(x)

* Cumulative Distribution Function (CDF) F(x)

* Mean, Variance, and Standard Deviation

Calculates moment number t using the moment generating function

What is the probability of drawing an ace from a deck of 52 cards?

What is the probability of drawing an ace from a deck of 52 cards?
With 4 Aces in the deck, the probability we draw an Ace is:
4/52
Simplifying this fraction, we get [B]1/13[/B]

What is the probability of picking a P from the word hippopotamus?

What is the probability of picking a P from the word hippopotamus?
There are 3 P's in a word with 12 characters.
So our probability is 3/12.
[URL='https://www.mathcelebrity.com/fraction.php?frac1=3%2F12&frac2=3%2F8&pl=Simplify']Using our fraction simplify calculator[/URL], this simplifies to 1/4.
Written as a decimal, it's 0.25.
So our answer is [B]1/4 or 0.25[/B]

What is the probability of rolling 12, 5 times in a row?

The only way you can roll a 12 with two dice is 6 and 6. Since each die roll is independent, we have:
[LIST]
[*]P(12) = P(6) * P(6)
[*]P(12) = 1/6 * 1/6
[*]P(12) = 1/36.
[/LIST]
Now, what is the probability we roll a 12 five times in a row? The same rules apply, each new roll is independent of the last, so we multiply:
[LIST]
[*]P(12, 12, 12, 12, 12) = 1/36 * 1/36 * 1/36 * 1/36 * /36
[*]P(12, 12, 12, 12, 12) = [B]1/60,466,176[/B] or [B]1.65381717e-8[/B]
[/LIST]

What is the probability that a month chosen at random has less than 31 days?

What is the probability that a month chosen at random has less than 31 days?
Months with 31 days:
[LIST=1]
[*]January
[*]March
[*]May
[*]July
[*]August
[*]October
[*]December
[/LIST]
7 months out of 12 have 31 days, so our probability is [B]7/12[/B]

what is the probabilty of tossing two coins and both landing on heads

what is the probabilty of tossing two coins and both landing on heads
We want P(HH). We type in [URL='https://www.mathcelebrity.com/cointoss.php?hts=HH&hct=+2&tct=+1&fct=+5>=no+more+than&nmnl=+2&htpick=tails&calc=1&montect=500&pl=Calculate+Probability']HH into our search engine[/URL] and we get:
P(HH) = [B]0.25 or 1/4[/B]

what’s the probability of rolling a 5 and then rolling a number less then 2

what’s the probability of rolling a 5 and then rolling a number less then 2
[U]Roll a 5:[/U]
There's only one 5 on a six sided die
P(X = 5) = 1/6
A number less than 2 is only 1:
P(X < 2) = P(X = 1)
P(X = 1) = 1/6
Since each event is independent, we multiply:
P(X = 5) * P(X = 1) = 1/6 * 1/6
P(X = 5) * P(X = 1) = [B]1/36[/B]

When five people are playing a game called hearts, each person is dealt ten cards and the two remain

When five people are playing a game called hearts, each person is dealt ten cards and the two remaining cards are put face down on a table. Because of the rules of the game, it is very important to know the probability of either of the two cards being a heart. What is the probability that at least one card is a heart?
Probability that first card is not a heart is 3/4 since 4 suits in the deck, hearts are 1/4 of the deck.
Since we don't replace cards, the probability of the next card drawn without a heart is (13*3 - 1)/51 = 38/51
Probability of both cards not being hearts is found by multiplying both individual probabilities:
3/4 * 38/51 = 114/204
Having at least one heart is found by subtracting this from 1 which is 204/204:
204/204 - 114/204 = 90/204
[URL='https://www.mathcelebrity.com/search.php?q=90%2F204&x=0&y=0']This reduces to[/URL] [B]15/34[/B]

Which of the following is NOT TRUE about the distribution for averages?

Which of the following is NOT TRUE about the distribution for averages?
a. The mean, median, and mode are equal.
b. The area under the curve is one.
c. The curve never touches the x-axis.
d. The curve is skewed to the right.
Answer is d, the curve is skewed to the right
For a normal distribution:
[LIST]
[*] The area under the curve for a standard normal distribution equals 1
[*] Mean media mode are equal
[*] Never touches the x-axis since in theory, all events have some probability of occuring
[/LIST]

Which of the following is the probability that a green marble will be selected from a bag containing

Which of the following is the probability that a green marble will be selected from a bag containing 9 red marbles 6 blue marbles 7 green marbles and 11 yellow marbles if one is selected randomly?
Total marbles in the bag:
9 red + 6 blue + 7 green + 11 yellow = 33
P(Green) = Green Marbles / Total Marbles
P(Green) = [B]7/33[/B]

Which of the following is the probability that subjects do not have the disease, but the test result

Which of the following is the probability that subjects do not have the disease, but the test result is positive?
a. Miss rate
b. False positive rate
c. Base rate
d. Disease rate
[B]b. [URL='http://sites.stat.psu.edu/~lsimon/stat250/sp00/solutions/misc/diagtests.htm']False positive rate[/URL][/B]

Which of the followings is the definition of power? a. Power is the probability of rejecting a null

Which of the followings is the definition of power?
a. Power is the probability of rejecting a null hypothesis
b. Power is the probability of accepting a null hypothesis
c. Power is the probability of accepting a false null hypothesis
d. Power is the probability of rejecting a false null hypothesis
[B]d. Power is the probability of rejecting a false null hypothesis[/B]

Yahtzee-1st Roll

Free Yahtzee-1st Roll Calculator - Calculates the probability of various scoring hands in the game of Yahtzee on the 1st roll of the dice.

You are using a spinner with the numbers 1-10 on it. Find the probability that the pointer will sto

You are using a spinner with the numbers 1-10 on it. Find the probability that the pointer will stop on an odd number or a number less than 4.
We want P(odd number) or P(n<4).
[LIST]
[*]Odd numbers are {1, 3, 5, 7, 9}
[*]n < 4 is {1, 2, 3}
[/LIST]
We want the union of these 2 sets:
{1, 2, 3, 5, 7, 9}
We have 6 possible pointers in a set of 10.
[B]6/10 = 3/5 = 0.6 or 60%[/B]

You choose an alpha level of .01 and then analyze your data.

(a) What is the probability that

You choose an alpha level of .01 and then analyze your data.
(a) What is the probability that you will make a Type I error given that the null hypothesis is true?
(b) What is the probability that you will make a Type I error given that the null hypothesis is false.
[B](a) 0.01. Instead, α is the probability of a Type I error given that the null hypothesis is true. If the null hypothesis is false, then it is impossible to make a Type I error.[/B]
[B](b) Impossible Instead, α is the probability of a Type I error given that the null hypothesis is true. If the null hypothesis is false, then it is impossible to make a Type I error.[/B]

you draw a card at random from a deck that contains 3 black cards and 7 red cards what is the probab

you draw a card at random from a deck that contains 3 black cards and 7 red cards what is the probability of you drawing a black card
Total cards = 3 black + 7 red
Total cards = 10
P(Black) = Black cards / Total Cards
P(Black) = [B]3/10 or 0.3[/B]

You roll two six-sided dice. What is the probability that the sum is less than 13?

You roll two six-sided dice. What is the probability that the sum is less than 13?
The probability is [B]1, or 100%[/B], since the maximum sum of two six-sided dice is 12.

You throw two dice. The red dice is fair but on the blue dice the probability of a 1=15%, probabilit

You throw two dice. The red dice is fair but on the blue dice the probability of a 1=15%, probability of a 2 is 25%, and the probability of any other number is 15%. What is the probability of getting 4?
Possible Rolls with a sum of 4:
[LIST]
[*]R = 1, B = 3
[*]R = 2, B = 2
[*]R = 3, B = 1
[/LIST]
Probabilities:
[LIST]
[*]R = 1, B = 3 = 1/6 * 15/100 = 15/600 = 1/40 = 0.025
[*]R = 2, B = 2 = 1/6 * 25/100 = 25/600 = 1/24 = 0.041667
[*]R = 3, B = 1 = 1/6 * 15/100= 15/600 = 1/40 = 0.025
[/LIST]
We add all three probabilities up to get:
0.025 + 0.025 + 0.014667 = [B]0.09166667[/B]

Z Score Lookup

Free Z Score Lookup Calculator - Given a Z-score probability statement from the list below, this will determine the probability using the normal distribution z-table.

* P(z < a)

* P(z <= a)

* P(z > a)

* P(z >= a)

* P(a < z < b) Calculates z score probability

* P(z < a)

* P(z <= a)

* P(z > a)

* P(z >= a)

* P(a < z < b) Calculates z score probability

___is the probability of a Type II error; and ___ is the probability of correctly rejecting a false

___is the probability of a Type II error; and ___ is the probability of correctly rejecting a false null hypothesis.
a. 1 - β; β
b. β; 1 - β;
c. α; β;
d. β; α
[B]b. β; 1 - β;[/B]
[LIST]
[*]H0 is true = Correct Decision 1 - α Confidence Level = Size of a Test α = Type I Error
[*]Ho is false = Type II Error β = Correct Decision 1 - β = Power of a Test
[/LIST]