event  89 results

event - a set of outcomes of an experiment to which a probability is assigned.

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

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 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

3 coins fall out of your pocket. What is the probability that all 3 will land tails up?
Since each event is independent, we have: P(T) * P(T) * P(T) 1/2 * 1/2 * 1/2 [B]1/8[/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]

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 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 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 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 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 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 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 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 catering service offers 3 appetizers, 6 main courses, and 4 desserts. A customer is to select 2 ap
A catering service offers 3 appetizers, 6 main courses, and 4 desserts. A customer is to select 2 appetizers, 3 main courses, and 3 desserts for a banquet. In how many ways can this be done? We use the combinations formula, and since each event is independent of the others, we multiply: 2 appetizers, 3 main courses, and 3 desserts = [URL='https://www.mathcelebrity.com/permutation.php?num=3&den=2&pl=Combinations']3C2[/URL] * [URL='https://www.mathcelebrity.com/permutation.php?num=6&den=3&pl=Combinations']6C3[/URL] * [URL='https://www.mathcelebrity.com/permutation.php?num=4&den=3&pl=Combinations']4C3[/URL] 2 appetizers, 3 main courses, and 3 desserts = 3 * 20 * 4 2 appetizers, 3 main courses, and 3 desserts = [B]240[/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 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 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 committee of 6 students are being selected from a class of 10 girls and 8 boys. How many committee
A committee of 6 students are being selected from a class of 10 girls and 8 boys. How many committees are possible if three must be girls and 3 must be boys? We want combinations. How many ways can we choose 3 boys from 8 boys: [URL='https://www.mathcelebrity.com/permutation.php?num=8&den=3&pl=Combinations']8 choose 3[/URL] = 56 We want combinations. How many ways can we choose 3 girls from 10 girls: [URL='https://www.mathcelebrity.com/permutation.php?num=10&den=3&pl=Combinations']10 choose 3[/URL] = 120 Our total choices are found by multiplying each event: Total committees = (8 boys choose 3) * (10 girls choose 3) Total committees = 56 * 120 Total committees = [B]6,720[/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 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 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 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 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 members-only speaker series allows people to join for \$16 and then pay \$1 for every event attended
A members-only speaker series allows people to join for \$16 and then pay \$1 for every event attended. What is the maximum number of events someone can attend for a total cost of \$47? Subtract the join fee from the total cost: \$47 - \$16 = \$31 Now divide this number by the cost per event: \$31 / \$1 = [B]31 events[/B]

A non-profit organization is having a couple’s banquet for a fundraiser. The banquet hall will only
A non-profit organization is having a couple’s banquet for a fundraiser. The banquet hall will only hold 250 people. The President, Vice-President, two volunteers, and a guest speaker will be working the event. How many couples will be able to attend the banquet? We subtract the 5 people working the event to get: 250 - 5 = 245 A couple is 2 people, so we have 245/2 = 122.5 We round down to [B]122 couples[/B].

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

a paper boy delivers thirteen paper to an apartment complex. if these deliveries compose one-seventh
a paper boy delivers thirteen paper to an apartment complex. if these deliveries compose one-seventh of his route, how many papers does he deliver Let d be the total number of deliveries the paper boy makes on the route. d We're given, d/7 = 13 d = 13 * 7 d = [B]91 [MEDIA=youtube]HRviz-3fn5c[/MEDIA][/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 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]

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]

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]

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]

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

Benny bought 8 new baseball trading cards to add to his collection. The next day his dog ate half of
Benny bought 8 new baseball trading cards to add to his collection. The next day his dog ate half of his collection. There are now only 47 cards left. How many cards did Benny start with? Let b be the number of baseball trading cards Benny started with. We have the following events: [LIST=1] [*]Benny buys 8 new cards, so we add 8 to get b + 8 [*]The dog ate half of his cards the next day, so Benny has (b + 8)/2 [*]We're told he has 47 cards left, so we set (b + 8)/2 equal to 47 [/LIST] (b + 8)/2 = 47 [B][U]Cross multiply:[/U][/B] b + 8 = 47 * 2 b + 8 = 94 [URL='https://www.mathcelebrity.com/1unk.php?num=b%2B8%3D94&pl=Solve']Type this equation into the search engine[/URL], we get [B]b = 86[/B].

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

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]

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]

Edna plans to treat her boyfriend Curt to dinner for his birthday. The costs of their date options a
Edna plans to treat her boyfriend Curt to dinner for his birthday. The costs of their date options are listed next to each possible choice. Edna plans to allow Curt to choose whether they will eat Mexican food (\$25), Chinese food (\$15), or Italian food (\$30). Next, they will go bowling (\$20), go to the movies (\$30) or go to a museum (\$10). Edna also is deciding between a new wallet (\$12) and a cell phone case (\$20) as possible gift options for Curt. What is the maximum cost of this date? Edna has 3 phases of the date: [LIST=1] [*]Dinner [*]Event after dinner [*]Gift Option [/LIST] In order to calculate the maximum cost of the date, we take the maximum cost option of all 3 date phases: [LIST=1] [*]Dinner - Max price is Italian food at \$30 [*]Event after dinner - Max price is movies at \$30 [*]Gift Option - Max price option is the cell phone cast at \$20 [/LIST] Add all those up, we get: \$30 + \$30 + \$20 = [B]\$80[/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]

Ethan has \$9079 in his retirement account, and Kurt has \$9259 in his. Ethan is adding \$19per day, wh
Ethan has \$9079 in his retirement account, and Kurt has \$9259 in his. Ethan is adding \$19per day, whereas Kurt is contributing \$1 per day. Eventually, the two accounts will contain the same amount. What balance will each account have? How long will that take? Set up account equations A(d) where d is the number of days since time 0 for each account. Ethan A(d): 9079 + 19d Kurt A(d): 9259 + d The problems asks for when they are equal, and how much money they have in them. So set each account equation equal to each other: 9079 + 19d = 9259 + d [URL='https://www.mathcelebrity.com/1unk.php?num=9079%2B19d%3D9259%2Bd&pl=Solve']Typing this equation into our search engine[/URL], we get [B]d = 10[/B]. So in 10 days, both accounts will have equal amounts in them. Now, pick one of the account equations, either Ethan or Kurt, and plug in d = 10. Let's choose Kurt's since we have a simpler equation: A(10) = 9259 + 10 A(10) = \$[B]9,269 [/B] After 10 days, both accounts have \$9,269 in them.

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

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]

Geocache puzzle help
In the first hour, he sold one-half of his sticks, plus one-half of a stick. The next hour, he sold one-third of his remaining sticks plus one-third of a stick. In the third hour, he sold one-fourth of what he had left, plus three-fourths of a stick. The last hour, he sold one-fifth of the remaining sticks, plus one-fifth of a stick. He did not cut up any sticks to make these sales. He returned home with 19 sticks. How many did he originally take to the event?

Geocache puzzle help
Ok. To go further in this equation. It reads: ...How many did he originally take to the event? Multiply the answer by 3 and reverse the digits. This will give you the answer for ACH in the coordinates. Does that make sense to reverse 303? :-/ Thank you for your help!!

Geocache puzzle help
Let me post the whole equation paragraph: Brainteaser # 1: Answer for ACH A fellow geocacher decided that he would try to sell some hand-made walking sticks at the local geocaching picnic event. In the first hour, he sold one-half of his sticks, plus one-half of a stick. The next hour, he sold one-third of his remaining sticks plus one-third of a stick. In the third hour, he sold one-fourth of what he had left, plus three-fourths of a stick. The last hour, he sold one-fifth of the remaining sticks, plus one-fifth of a stick. He did not cut up any sticks to make these sales. He returned home with 19 sticks. How many did he originally take to the event? Multiply the answer by 3 and reverse the digits. This will give you the answer for ACH in the coordinates. Make sure to multiply and reverse the digits. What would the answer be?

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

If 13,754 people voted for a politician in his first election, 15,420 voted for him in his second el
If 13,754 people voted for a politician in his first election, 15,420 voted for him in his second election, and 8,032 voted for him in the first and second elections, how many people voted for this politician in the first or second election? Let P(A) be the first election votes, P(B) be the second election votes, and P(A ? B) be votes for both the first AND the second elections. We want P(A U B). Use our [URL='http://www.mathcelebrity.com/probunion2.php?pa=+13754&pb=15420&paintb=8032&aub=+&pl=Calculate']two event calculator[/URL] P(A U B) = P(A) + P(B) - P(A ? B) P(A U B) = 13,754 + 15,420 - 8032 P(A U B) = 29,174 - 8,032 P(A U B) = [B]21,142[/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 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 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]

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]

In rolling a die, the event E is getting a number greater than or equal to 3. What is the complement
In rolling a die, the event E is getting a number greater than or equal to 3. What is the complement of the event? The complement E' is everything but the event. So we have: E = P(n >= 3) E' = [B]P(n < 3)[/B]

In the movie Die Hard: With a Vengeance, in one of the action scenes, the characters Mc Clane and Ca
In the movie Die Hard: With a Vengeance, in one of the action scenes, the characters Mc Clane and Carver were caught in a breathtaking scenario where they need to keep a bomb from exploding, and the only way to prevent explosion is to put exactly four gallons of water on a scale. How would they do it if they only have a five - gallon and a three gallon jug? [LIST=1] [*]Fill the 5-gallon jug all the way. [*]Pour water into the 3 gallon jug until it is full. [*]Now you have 2 gallons in the 5-gallon jug and a full 3 gallons in the 3-gallon jug. [*]Empty the 3-gallon jug. [*]Pour the 2 gallons of water still in the 5-gallon jug into the 3-gallon jug. [*]Now the 3-gallon jug has 2 gallons of water in it, and 1 gallon of empty space. [*]Fill up the 5 gallon jug all the way, and then pour water out of the 5-gallon jug into the 3-gallon jug until the 3-gallon jug is full. [*]This leaves exactly [B]4 gallons[/B] in the 5-gallon jug. [/LIST]

juan sells raffle tickets at a charity event for \$6 each.How many tickets does he have to sell to ma
juan sells raffle tickets at a charity event for \$6 each.How many tickets does he have to sell to make \$144? Tickets needed = Total Sales / Cost per ticket Tickets needed = 144/6 Tickets needed = [B]24[/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]

Multinomial Distribution
Free Multinomial Distribution Calculator - Given a set of xi counts and a respective set of probabilities θi, this calculates the probability of those events occurring.

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

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

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)

Prove P(A’) = 1 - P(A)
Prove P(A’) = 1 - P(A) The sample space S contains an Event A and everything not A, called A' We know P(S) = 1 P(S) = P(A U A') P(A U A') = 1 P(A) + P(A') = 1 subtract P(A) from each side: P(A’) = 1 - P(A) [MEDIA=youtube]dNLl_8vejyE[/MEDIA]

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]

Serial numbers for a product are to be using 3 letters followed by 4 digits. The letters are to be t
Serial numbers for a product are to be using 3 letters followed by 4 digits. The letters are to be taken from the first 8 letters of the alphabet with no repeats. The digits are taken from numbers 0-9 with no repeats. How many serial numbers can be generated The serial number is organized with letters (L) and digits (D) like this: LLLDDDD Here's how we get the serial number: [LIST=1] [*]The first letter can be any of 8 letters A-H [*]The second letter can be any 7 of 8 letters A-H [*]The third letter can be any 6 of 8 letters A-H [*]The fourth digit can be any of 10 digits 0-9 [*]The fifth digit can be any 9 of 10 digits 0-9 [*]The sixth digit can be any 8 of 10 digits 0-9 [*]The seventh digit can be any 7 of 10 digits 0-9 [/LIST] We multiply all possibilities: 8 * 7 * 6 * 10 * 9 * 8 * 7 [B]1,693,440[/B]

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]

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 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 sample space for a coin being tossed twice
the sample space for a coin being tossed twice Since each toss results in 2 outcomes, we have 2^2 = 4 possible events in the sample space: [LIST=1] [*]H,H [*]H,T [*]T,H [*]T,T [/LIST]

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 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 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]

There were 175 tickets sold for the upcoming event in the gym. If students tickets cost \$5 and adult
There were 175 tickets sold for the upcoming event in the gym. If students tickets cost \$5 and adult tickets are \$8, tell me how many tickets were sold if gate receipts totaled \$1028? Let s be the number of student tickets and a be the number of adult tickets. We are given: a + s = 175 8a + 5s = 1028 There are 3 ways to solve this, all of which give us: [B]a = 51 s = 124 [/B] [URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+s+%3D+175&term2=8a+%2B+5s+%3D+1028&pl=Substitution']Substitution Method[/URL] [URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+s+%3D+175&term2=8a+%2B+5s+%3D+1028&pl=Elimination']Elimination Method[/URL] [URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+s+%3D+175&term2=8a+%2B+5s+%3D+1028&pl=Cramers+Method']Cramers Method[/URL]

Time Weighted Interest Method
Free Time Weighted Interest Method Calculator - Solves for Interest Rate based on 2 annual asset value events other than beginning or ending value using the Time Weighted Method

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 Ac)=1, where Ac is the complement of A. (d) In a hypothesis testing, if the p-value is less than the significance level ?, 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. Enter the size of the set that corresponds to the event that both dice are odd.
Two dice are rolled. Enter the size of the set that corresponds to the event that both dice are odd. If dice 1 is odd, then we have the following face values: {1, 3, 5} If dice 2 is odd, then we have the following face values: {1, 3, 5} [URL='https://www.mathcelebrity.com/2dice.php?gl=1&opdice=1&pl=Both+Odd&rolist=+2%2C3%2C9%2C10&dby=2%2C3%2C5&ndby=4%2C5&montect=+100']From this 2 dice odds face link[/URL], we see that the size of the set is 9. [LIST=1] [*]{1, 1} [*]{1, 3} [*]{1, 5} [*]{3, 1} [*]{3, 3} [*]{3, 5} [*]{5, 1} [*]{5, 3} [*]{5, 5} [/LIST]

What is the average of 7 consecutive numbers if the smallest number is called n?
What is the average of 7 consecutive numbers if the smallest number is called n? [LIST] [*]First number = n [*]Second number = n + 1 [*]Third number = n + 2 [*]Fourth number = n + 3 [*]Fifth number = n + 4 [*]Sixth number = n + 5 [*]Seventh number = n + 6 [/LIST] Average = Sum of all numbers / Total numbers Average = (n + n + 1 + n + 2 + n + 3 + n + 4 + n + 5 + n + 6)/7 Average = 7n + 21/7 Factor out a 7 from the top: 7(n + 3)/7 Cancel the 7's: [B]n + 3[/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]

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]