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 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 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 company has 81 employees of whom x are members of a union how many are not in the union You can either be a union member or a non-union member. This is our sample space. If we have 81 employees and x are union members, this means that: Non-Union membes = [B]81 - x[/B]

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 six-sided die is rolled. Describe the sample space. [B]{1, 2, 3, 4, 5, 6}[/B]

A family has 4 children. Give the sample space in regards to the genders of the children. Children can either be male or female. Therefore, the sample space is 2 * 2 * 2 * 2 = 16 possible combinations. [LIST=1] [*]MMMM [*]MMMF [*]MMFM [*]MFMM [*]FMMM [*]MMFF [*]MFFM [*]FFMM [*]MFMF [*]FMFM [*]MMMF [*]FMMM [*]FFFM [*]MFFF [*]FMMF [*]FFFF [/LIST] [MEDIA=youtube]W0bthXg-368[/MEDIA]

A pair of standard dice is rolled, how many possible outcomes are there? We want the number of outcomes in the sample space. The first die has 6 possibilities 1-6. The second die has 6 possibilities 1-6. Our sample space count is 6 x 6 = [B]36 different outcomes [/B] [LIST=1] [*](1, 1) [*](1, 2) [*](1, 3) [*](1, 4) [*](1, 5) [*](1, 6) [*](2, 1) [*](2, 2) [*](2, 3) [*](2, 4) [*](2, 5) [*](2, 6) [*](3, 1) [*](3, 2) [*](3, 3) [*](3, 4) [*](3, 5) [*](3, 6) [*](4, 1) [*](4, 2) [*](4, 3) [*](4, 4) [*](4, 5) [*](4, 6) [*](5, 1) [*](5, 2) [*](5, 3) [*](5, 4) [*](5, 5) [*](5, 6) [*](6, 1) [*](6, 2) [*](6, 3) [*](6, 4) [*](6, 5) [*](6, 6) [/LIST]

An airplane carries 500 passengers 45% are men, 20% are children. The number of women in the airplane is If we assume the sample space is either men, women, or children to get 100% of the passengers, we have: PercentWomen = 100% - Men - Children PercentWomen = 100% - 45% - 20% PercentWomen = 35% Calculate Women passengers Women passengers = Total passengers * Percent Women Women passengers = 500 * 35% Women passengers = [B]175[/B]

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]

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]

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 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]

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).

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) 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 for flipping a coin 3 times Each flip gives us 2 possible outcomes, heads or tails. So we have: 2 * 2 * 2 = 8 possible outcomes [LIST=1] [*]HHH [*]HHT [*]HTH [*]HTT [*]THH [*]THT [*]TTH [*]TTT [/LIST]

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

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 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]

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 ?, 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]

What is the sample space for a 10 sided die? Sample space means the set of all possible outcomes. For a 10-sided die, we have: [B]{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}[/B]

Write a sample space for rolling a dice twice Each die roll has 6 possible outcomes. So 2 die-rolls has 6^2 = 36 possible outcomes: [LIST=1] [*]1,1 [*]1,2 [*]1,3 [*]1,4 [*]1,5 [*]1,6 [*]2,1 [*]2,2 [*]2,3 [*]2,4 [*]2,5 [*]2,6 [*]3,1 [*]3,2 [*]3,3 [*]3,4 [*]3,5 [*]3,6 [*]4,1 [*]4,2 [*]4,3 [*]4,4 [*]4,5 [*]4,6 [*]5,1 [*]5,2 [*]5,3 [*]5,4 [*]5,5 [*]5,6 [*]6,1 [*]6,2 [*]6,3 [*]6,4 [*]6,5 [*]6,6 [/LIST]

You roll a red die and a green die. What is the size of the sample space of all possible outcomes of rolling these two dice, given that the red die shows an even number and the green die shows an odd number greater than 1? [LIST] [*]Red Die Sample Space {2, 4, 6} [*]Green Die Sample Space {3, 5} [*]Total Sample Space {(2, 3), (2, 5), (4, 3), (4, 5), (6, 3), (6, 5)} [*]The sie of this is 6 elements. [/LIST]

You roll a standard, fair, 5-sided die and see what number you get. Find the sample space of this experiment. Write your answer using { } symbols, and write your values in order with a comma but no spaces between Sample Space: [B]{1,2,3,4,5}[/B]