knewton alta chapter 3 part 2

Brian has two football games to play in this week. Let A represent the event that his team loses the first game and B represent the event that his team loses the second game. If A and B are independent events with P(A)=0.2 and P(B)=0.5, find P(A AND B). Give your answer as a percent, rounded to two decimal places if necessary.

$10\%$10%​ Remember that for independent events, P(A AND B)=P(A)P(B) So, plugging in the values we are given, we find thatP(A AND B)=(0.20)(0.50)=0.1 The probability Brian's team loses both the first and second game is 0.1×100=10%.

If A and B are events with P(A)=0.9, P(B)=0.3, P(A OR B)=0.95, find P(A AND B).

$0.25$0.25​ First, it is helpful to write down the addition rule for probabilities: P(A OR B)=P(A)+P(B)−P(A AND B) Now, rearranging this, we find thatP(A AND B)=P(A)+P(B)−P(A OR B)Plugging in the known values, we findP(A AND B)=0.9+0.3−0.95=0.25

The probability that a student is a business major and in a statistics course is 0.29. The probability that a randomly chosen student is in a statistics course given that the student is a business major is 0.67. What is the probability a student is a business major? Round your answer to three decimal places.

Correct answers:$0.433$0.433​ Remember the multiplication rule for conditional probability: P(A AND B)=P(A|B)P(B) Rearranging, we find thatP(B)=P(A AND B)P(A|B) So if we think of A as being the event that a student is in a statistics course and B as being the event a student is a business major, then we can plug in the known information to find P(B)=0.29/0.67≈0.433

If A and B are events with P(A)=0.4, P(A OR B)=0.89, P(A AND B)=0.01, find P(B).

Correct answers:$0.5$0.5​ First, it is helpful to write down the addition rule for probabilities: P(A OR B)=P(A)+P(B)−P(A AND B) Now, rearranging this, we find thatP(B)=P(A OR B)+P(A AND B)−P(A)Plugging in the known values, we findP(B)=0.89+0.01−0.4=0.5

If A and B are independent events with P(A)=0.20 and P(A AND B)=0.12, find P(B). Give your answer as a percentage, rounded to two decimal places if necessary.

Correct answers:$60\%$60%​ Remember that for independent events, P(A AND B)=P(A)P(B) Solving for P(B), we find thatP(B)=P(A AND B)P(A)Now, plugging in the values we are given, we find thatP(B)=0.120.20=0.6 To change the decimal to a percent, we multiply by 100: 0.6×100=60%

A group of 120 students at a high school were asked if they prefer streaming to live TV. The results are shown in the table above. MaleFemaleTotalStreaming186785Live TV53035Total2397120 Given that a randomly selected survey participant is a male, what is the probability that this student prefers live TV? Give your answer as a fraction in simplest form.

Correct answers:$\frac{5}{23}$523​​ Because our given information is that the chosen participant is male, we want to focus on that column of the table. MaleFemaleTotalStreaming186785Live TV53035Total2397120 The total number of those that are males is 23. Of these, 5 prefer live TV to streaming. Therefore, the probability that a randomly selected male student that was surveyed prefers live TV is 523.

A group of 155 students at a private university were asked if they are full-time or part-time and if they own a car. The following table shows the results.

Correct answers:$\frac{7}{29}$729​​ Because our given information is that the is Part-Time, we want to focus on that column of the table. In that row, there are 29 students total, and 7 of them did not own a car, therefore the probability is 729.

How To Addition Rule for Probabilities

If A and B are events defined on a sample space, then P(A OR B)=P(A)+P(B)−P(A AND B)

Multiplication Rule for Independent Events

If A and B are independent events, then P(A AND B)=P(A)P(B)

If A and B are events with P(A)=0.5, P(A OR B)=0.65, P(A AND B)=0.15, find P(B).

Correct answers:$0.3$0.3​ First, it is helpful to write down the addition rule for probabilities: P(A OR B)=P(A)+P(B)−P(A AND B) Now, rearranging this, we find thatP(B)=P(A OR B)+P(A AND B)−P(A)Plugging in the known values, we findP(B)=0.65+0.15−0.5=0.3

The Multiplication Rule for Conditional Probability:

The probability of A given B is the probability of A AND B divided by the probability of B. P(A|B)=P(A AND B)/P(B)

Conditional Probability

The probability of one event based on the probability of another event already happening.

If A and B are independent events with P(A)=0.90 and P(A AND B)=0.54, find P(B). Give your answer as a decimal rounded to two decimal places.

Correct answers:$0.60$0.60​ Remember that for independent events, P(A AND B)=P(A)P(B) Solving for P(B), we find thatP(B)=P(A AND B)/P(A)Now, plugging in the values we are given, we find thatP(B)=0.54/0.90=0.60

without replacement

When a population element can be selected only one time, so after choosing an element it is set aside and is no longer available to be chosen again

If A and B are events with P(A)=0.7, P(B)=0.2, P(A AND B)=0.17, find P(A OR B).

$0.73$0.73​ First, it is helpful to write down the addition rule for probabilities: P(A OR B)=P(A)+P(B)−P(A AND B) Now, plugging in the known values, we find P(A OR B) =0.7+0.2−0.17=0.73

If you roll a fair die and then flip a fair coin, what is the probability that you roll greater than 1 and get heads with the coin? Write your answer as a fraction.

5/12

Given that the probability of a student eating at the cafeteria and a student living off campus is 0.07, and the probability of a student eating at the cafeteria given that the student lives off campus is 0.20, what is the probability of a student living off campus?

$0.35$0.35​ Remember the multiplication rule for conditional probability: P(B AND A)=P(B|A)P(A) Rearranging, we find thatP(A)=P(B AND A)P(B|A) So if we think of A as being the event a student lives off campus and B as being the event a student eats at the cafeteria, then we can plug in the known information to find P(A)=0.070.20=0.35

If A and B are events with P(A)=0.6 P(A OR B)=0.98 P(A AND B)=0.02, find P(B).

$0.4$0.4​ First, it is helpful to write down the addition rule for probabilities: P(A OR B)=P(A)+P(B)−P(A AND B) Now, rearranging this, we find that P(B)=P(A OR B)+P(A AND B)−P(A) Plugging in the known values, we find P(B)=0.98+0.02−0.6=0.4

If A and B are events with P(A)=0.8, P(A OR B)=0.87, P(A AND B)=0.23, find P(B).

.3

The results of an experiment are shown below. Calculate E, the expected value of the experiment. Write your answer as a decimal.

Correct answers:$6.4$6.4​ E=2⋅1/10+4⋅3/10+6⋅2/10+8⋅1/10+10⋅3/10=6.4.

Given that P(B|A)=0.84 and P(A)=0.43, what is P(B AND A)?

Correct answers:$0.3612$0.3612​ Remember the multiplication rule for conditional probability: P(B AND A)=P(B|A)P(A) So plugging in the values that we know, we findP(B AND A)=(0.84)(0.43)=0.3612

The probability that an adult is actively looking for a job is 0.18. The probability of an adult being unemployed and actively looking for a job is 0.04. What is the probability of an adult being unemployed given that they are actively looking for a job? Round your answer to two decimal places.

$0.22$0.22​ Remember the multiplication rule for conditional probability: P(B AND A)=P(B|A)P(A) Rearranging, we see thatP(B|A)=P(B AND A)P(A) So if we think of A as being the event that an adult is actively looking for a job and B as the event that an adult is unemployed, then we can plug in the known information to find P(B|A)=0.040.18≈0.22

If the probability of a student spending time reading is 0.70, the probability of spending time watching TV is 0.10, and the probability of spending time reading and watching TV is 0.01, what is the probability of a student spending time reading or watching TV?

$0.79$0.79​ Let A be the event spending time reading and B be the event spending time watching TV. It is helpful to write down the addition rule for probabilities: P(A OR B)=P(A)+P(B)−P(A AND B) Now, plugging in the known values, we findP(A OR B)=0.7+0.1−0.01=0.79So the probability of spending time reading or watching TV is 0.79.

The probability that an adult visits their dentist annually is 0.58. The probability that an adult has mouth pain given that they visit their dentist annually is 0.34. What is the probability that an adult has mouth pain and visits their dentist annually? Round your answer to three decimal places.

.197

A community festival organizes a fundraiser for their local park. They sold 25,000 tickets at $20 each. The grand prize is $5,000. There are two prizes valued at $2,000, three prizes valued at $1,000, five prizes of $500, and ten prizes of $100. What is the expected value of this raffle? Round to the nearest cent. Do not round until the final calculation.

A community festival organizes a fundraiser for their local park. They sold 25,000 tickets at $20 each. The grand prize is $5,000. There are two prizes valued at $2,000, three prizes valued at $1,000, five prizes of $500, and ten prizes of $100. What is the expected value of this raffle? Round to the nearest cent. Do not round until the final calculation.

There are two known potential defects on a certain brand of computer. Let A be the event that the keyboard malfunctions and B be the event that certain types of memory will randomly erase. The probability of events A and B happening is 0.54. If A and B are independent events and P(A)=0.90, what is P(B)?

Solution: Since A and B are independent, we know that P(A AND B)=P(A)⋅P(B) Rearranging, we can solve for P(B): P(A AND B)P(A)0.540.90.6=P(B)=P(B)=P(B) That is, the probability that the memory will randomly erase is 0.6.

The probability that a person uses public transit given that they also own a car is 8/14. If we know that 32 people in a sample use public transit and 14 people own cars, fill in the Venn diagram below with the number of people to reflect this probability. Let Event A represent the people who use public transit, and Event B represent the people who own cars.

We are given that P(A|B)=814, so we know that there are 8 people in A AND B, so we fill in 8 for Response 2. We are given that there are 32 people total who use public transit, so Response 1 is 32−8=24. We are given that there are 14 people total who own cars, so Response 3 is 14−8=6.

A company is planning its spring advertising campaign. Its marketing team is planning to mail out different types of advertisements, at random, to their town's residents each week. They have the following types of advertisements ready: Week 1Week 2Week 3Small PosterBrochureLeafletPostcardBrochureBusiness CardPostcardSmall Poster How many different possible ways can a resident receive an advertisement?

We could draw a tree diagram here to see all the possible outcomes, or we can use the Fundamental Counting Principle and quickly calculate the number of outcomes. The Fundamental Counting Principle tells us that we multiple the number of choices each week by one another to get the total number of possibilities. Possibilities=Week 1×Week 2×Week 3 Possibilities=2×3×3=18

with replacement

When a population element can be selected more than one time, so after choosing an element it is put back in before the next one is taken.

Example 4 Question: A bag of candy contains 3 blue candies, 4 green candies, and 6 red candies. If you draw two candies, one at a time and without replacement, what is the probability that you will draw a green candy and a blue candy?

Solution: Let G be the event of drawing a green candy on the first draw and B be the event of drawing a blue candy on the second draw. Then P(B|G) is the probability of drawing a blue candy given that the first candy was green, and P(B AND G) is the probability of drawing a green and a blue candy. P(G)=4/13; P(B|G)=3/12=14. Notice that here there are 3 possible blue candies and 12 candies left after drawing the first green candy (since we drew without replacement). We want to know P(B AND G), the probability of drawing a green and a blue candy. The Multiplication Rule for Conditional Probabilities gives that: P(B|G)=P(B AND G)P(G) Rearranging using algebra, we have: P(B AND G)=P(B|G)⋅P(G) Plugging in the probabilities from above, we have P(B AND G)=1/4⋅4/13=1/13

The probability that a debt holder has student loan debt, given that they also have credit card debt is 1130. If we know that 13 people in a sample of debt holders have student loan debt and 30 people have credit card debt, fill in the Venn diagram below with the number of debt holders to reflect this probability. Let Event A represent the people with student loans, and Event B represent the people with credit card debt.

We are given that P(A|B)=1130, so we know that there are 11 debt holders in A AND B, so we fill in 11 for Response 2. We are given that there are 13 people total who have student loans, so Response 1 is 13−11=2. We are given that there are 30 people total who have credit card debt, so Response 3 is 30−11=19

If A and B are events with P(A)=0.5, P(B)=0.3, P(A OR B)=0.67, find P(A AND B).

Correct answers:$0.13$0.13​ First, it is helpful to write down the addition rule for probabilities: P(A OR B)=P(A)+P(B)−P(A AND B) Now, rearranging this, we find thatP(A AND B)=P(A)+P(B)−P(A OR B)Plugging in the known values, we findP(A AND B)=0.5+0.3−0.67=0.13

Fill in the following contingency table and find the number of students who both read mysteries AND read comics.

Correct answers:$13$13​ By using the known totals along the rows and columns you can fill in the rest of the contingency table. For example, looking at the row of totals in the table, we know that the unknown number plus 44 is 101, so the missing number must be 57. Continuing in this way, we can fill in the entire table: From this, we can see that the number of students who both read mysteries and read comics is 13.

Let S={1,2,3,4} be a sample space with P(x)=kx where x is a member of {1,3}, and P(x)=k(x−1) where x is a member of {2,4}, where k is a constant. Find E(S).

Correct answers:$3$3​ First we find k by summing the probabilities and equating to one, i.e. k+(2−1)k+3k+(4−1)k=1, which gives k=18. HenceE(S)=1⋅18+2⋅18+3⋅38+4⋅38=3.

Consider a set S={1,2,3,4,5} and P(x)=x−1/10, where x is a member of S. Calculate E(S).

Correct answers:$4$4​ Multiply each event by the probability of then event and then add up the products to find E(S). E(S)E(S)=1P(1)+2P(2)+3P(3)+4P(4)+5P(5) =1⋅0+2⋅1/10+3⋅2/10+4⋅3/10+5⋅4/10=4.

Let D be the event that a randomly chosen person has seen a dermatologist. Let S be the event that a randomly chosen person has had surgery for skin cancer. Identify the answer which expresses the following with correct notation: The probability that a randomly chosen person has had surgery for skin cancer, given that the person has seen a dermatologist.

P(S|D) Remember that in general, P(A|B) is read as "The probability of A given B". Here we are given that the person has seen a dermatologist, so the correct answer is P(S|D).

At a certain school, intro to economics and intro to calculus meet at the same time, so it is impossible for a student take both classes. If the probability that a student takes intro to economics is 0.57, and the probability that a student takes intro to calculus 0.17, what is the probability that a student takes intro to economics or into to calculus?

$0.74$0.74​ Because it is impossible for a student to take both intro to economics and intro to calculus, we see that they are mutually exclusive events. Therefore, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) So we find thatP(A OR B)=P(A)+P(B)=0.57+0.17=0.74 The probability that a student takes into to economics or intro to calculus is the sum of the individual probabilities, which is 0.74.

Given that P(B AND A)=0.03 and P(A)=0.11, what is P(B|A)? Give your answer as a percent. Round your answer to two decimal places.

$27.27\ \%$27.27 %​ Remember the multiplication rule for conditional probability: P(B AND A)=P(B|A)P(A) Rearranging, we see thatP(B|A)=P(B AND A)P(A)Now, plugging in the values we were given, we find thatP(B|A)=0.03/0.11≈0.2727 To rewrite this decimal as a percent, we multiply by 100: 0.2727×100=27.27%.

Darryl plays a card game with his sister. He uses a 52 card deck and she chooses 1 card. If she chooses the 4 of hearts, she will win $10 and if she chooses any spade, she loses $2. If any other card is drawn, they break even. What is the expected value for Darryl's sister? Round to the nearest cent. Do not round until your final calculation. Note: within a deck of cards, there are 4 equal suits: hearts, clubs, diamonds, and spades.

1$-0.31$−0.31​ Recall, the expected value is found by multiplying the value and the probability for that event. Here, Darryl's sister can win $10, lose $2, or break even. E(x)=10⋅152−2⋅1352+0⋅3852 E(x)=1052−2652+0 E(x)=−1652=−0.30769...=0.31 She can expect to lose $0.31 each time she chooses a card.

the probability of either of two events happening (or both happening) is equal to the probability of the first event happening plus the probability of the second event happening, minus the probability that both occur.

Addition rule for probabilities:

Conditional Probability: The probability of one event based on the probability of another event already happening. The Multiplication Rule for Conditional Probability: The probability of A given B is the probability of A AND B divided by the probability of B. P(A|B)=P(A AND B)P(B) With Replacement: When a population element can be selected more than one time, so after choosing an element it is put back in before the next one is taken. Without Replacement: When a population element can be selected only one time, so after choosing an element it is set aside and is no longer available to be chosen again.

Conditional Probability: The probability of one event based on the probability of another event already happening. The Multiplication Rule for Conditional Probability: The probability of A given B is the probability of A AND B divided by the probability of B. P(A|B)=P(A AND B)P(B) With Replacement: When a population element can be selected more than one time, so after choosing an element it is put back in before the next one is taken. Without Replacement: When a population element can be selected only one time, so after choosing an element it is set aside and is no longer available to be chosen again.

An experiment gives us the following results: Outcome−10−2015Probability1511011011012 What is the expected value of this experiment?

Solution By definition, the expected value of this experiment is E=−10⋅15−2⋅110+0⋅110+1⋅110+5⋅12=−105−210+110+52=−2010−210+110+2510=−20−2+1+2510=410=2/5.

Question The probability that a randomly chosen student plays in the band is 150. The probability that a randomly chosen student sings in the choir and plays in the band is 1200. What is the probability that a randomly chosen student sings in the choir, given that the student plays in the band?

Solution If B is the event that a random student plays in the band, and C is the event that a random student sings in the choir. Then we are told P(B)=150 and P(C AND B)=1200. We are asked for P(C|B). Using the formula, we find that P(C|B)=P(C AND B)P(B)=1/2001/50=50200=14 So the probability that a randomly chosen student sings in the choir given that he or she plays in the band is 1/4.

Venn Diagram

a picture that represents the outcomes of an experiment, generally consisting of a box that represents the sample space together with circles or ovals to represent events

The probability that a product passes quality-control inspections is 0.88. The probability that it passes quality-control inspections and is packaged for shipment immediately is 0.03. What is the probability that it is packaged for shipment immediately given that it passed quality-control inspections? Round your answer to three decimal places.

$0.034$0.034​ The probability that a product passes quality-control inspections is 0.88. The probability that it is packaged for shipment immediately and passes quality-control inspections is 0.03. What is the probability that it is packaged for shipment immediately given that it passed quality-control inspections? Remember the multiplication rule for conditional probability: P(B AND A)=P(B|A)P(A) Rearranging, we see thatP(B|A)=P(B AND A)P(A) So if we think of A as the product passing quality-control inspections and B as the event that the product is packaged to ship immediately, then we can plug in the known information to find P(B|A)=0.030.88≈0.034

Suppose A and B are mutually exclusive events, and that P(B)=0.24 and P(A OR B)=0.27. Find P(A).

Correct answers:$0.03$0.03​ Remember that when A and B are mutually exclusive events, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) Rearranging, we find thatP(A)=P(A OR B)−P(B)So plugging in the given values, we findP(A)=0.27−0.24=0.03

Given that P(A AND B)=0.70 and P(A|B)=0.94, what is P(B)? Round your answer to three decimal places.

Correct answers:$0.745$0.745​ Remember the multiplication rule for conditional probability: P(A AND B)=P(A|B)P(B) Rearranging, we find thatP(B)=P(A AND B)P(A|B)Plugging in the values we were given, we find thatP(B)=0.700.94≈0.745

A NPR/Robert Wood Johnson Foundation/Harvard T.H. Chan School of Public Health poll asked adults whether they participate in a sport. Given that the probability that a randomly chosen adult is a woman is 0.51, and the probability that a randomly chosen adult is a woman and plays a sport is 0.08, what is the probability of an adult playing a sport given that the adult is a woman? Round your answer to two decimal places.

Correct answers:1$0.16$0.16​ Let M be the event that a randomly chosen adult plays a sport, and let W be the event that a randomly chosen adult is a woman. Then according to the formula for conditional probability, P(M|W)=P(M AND W)P(W) So plugging in what we know, we find that P(M|W)=0.08/0.51=0.16 So the probability that a random adult plays a sport given that the adult is a woman is 0.16.

Multiplication Rule for Conditional Probability The conditional probability P(A|B) is given by the formula P(A|B)=P(A AND B)P(B) The probability P(A AND B) is given by the formulaP(A AND B)=P(A|B)P(B)

Multiplication Rule for Conditional Probability The conditional probability P(A|B) is given by the formula P(A|B)=P(A AND B)P(B) The probability P(A AND B) is given by the formulaP(A AND B)=P(A|B)P(B)

Question Let S={1,4,8,16,32,64} be a sample space. If P(1)=1/32 and P(2k)=1/2k−1 for each k from 2 to 6, find the expected value of the event E={4,8,16,32}.

Solution We know P(1)=1/32,P(4)=1/2, P(8)=1/4 and P(16)=1/8. Therefore the expected value of E is given by E(E)=4⋅12+8⋅14+16⋅18+32116=2+2+2+2=8.

Fill in the following contingency table and find the number of students who both do not have a cat AND do not have a dog.

34

A couple has two children. Let A be the event that their first child is a boy, and note that P(A)=51.2%. Let B be the event that their second child is a girl, with P(B)=48.8%. A and B are independent events. What is the probability that the couple has a first child that is a boy and a second child that is a girl? Give your answer as a decimal rounded to two decimal places.

Correct answers:$0.25$0.25​ Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=51.2%⋅48.8%=0.512⋅0.488≈0.25

independent events

Events are said to be independent if the probability of one occurring does not affect the probability of the other happening.

Given that the probability of a student spending time reading is 0.59, and the probability of a student doing well on an exam and spending time reading is 0.58, what is the probability of a student doing well on an exam given that the student spends time reading? Round your answer to three decimal places.

.983

A NPR/Robert Wood Johnson Foundation/Harvard T.H. Chan School of Public Health poll asked adults whether they participate in a sport. Given that the probability that a randomly chosen adult is a woman is 0.51, and the probability that a randomly chosen adult is a woman and plays a sport is 0.08, what is the probability of an adult playing a sport given that the adult is a woman? Round your answer to two decimal places.

1$0.16$0.16​ Let M be the event that a randomly chosen adult plays a sport, and let W be the event that a randomly chosen adult is a woman. Then according to the formula for conditional probability, P(M|W)=P(M AND W)P(W) So plugging in what we know, we find that P(M|W)=0.080.51=0.16 So the probability that a random adult plays a sport given that the adult is a woman is 0.16.

First year nursing students are randomly assigned to classes in a nursing school. Let A be the event that they are assigned to an Anatomy course and B be the event that they are assigned to a Psychology course. The probabilities of them being assigned to these courses are as follows: P(A)=23 and P(B)=25 If A and B are independent events, what is P(A AND B), the probability that a student will be assigned to both an Anatomy and a Psychology course? Give your answer as a fraction.

Correct answers:$\frac{4}{15}$415​​ Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=23⋅25=415

The probability that a randomly chosen college student smokes is 0.2. The probability that a randomly chosen college student is an athlete is 0.15. The probability that a randomly chosen college student smokes given that the student is an athlete is 0.1. What is the probability that a randomly chosen college student is a smoker and an athlete?

Let S be the event that a randomly chosen student smokes. Let A be the event that a randomly chosen student is an athlete. We are told P(S)=0.2, P(A)=0.15, and P(S|A)=0.1. We are asked to find P(S AND A). Using what we are told and the formula for the compound probability above, we find that P(S AND A)=P(S|A)P(A)=(0.1)⋅(0.15)=0.015 So the probability that a randomly chosen student is both an athlete and a smoker is0.015. Note that we did not use the information about P(S). Sometimes you may be given extraneous information in problems, so you need to be careful to apply the formula correctly.

Six friends at a university are taking classes in statistics or writing. Pamela and Sam are taking statistics; Nandini, Elliot, and Jill are taking writing; and Matt is taking both writing and statistics. This is shown in the Venn diagram below. Question: If we were to select one friend at random, what is the probability that they would be in either statistics or writing, but not both?

Solution: To solve this problem, it is pretty simple for us to count the number of friends that are not in both classes. There are 5 out of the 6 friends who are only in one of the courses, so

Question: A group of friends take classes at a university in biology, chemistry, and physics. If you selected a friend at random, the probabilities that they take certain classes is given below: The probability that they take biology class is P(B)=13; The probability that the friend takes chemistry is P(C)=12; A student takes physics with the probability P(P)=16; A student takes both biology and physics with a probability of 112. What's the probability that a randomly chosen friend takes biology or physics, P(B OR P)?

Solution: Without a diagram, or information about how many friends there are, we have to use the Addition Rule for Probabilities to P(B OR P): P(B OR P)=P(B)+P(P)−P(B AND P)=13+16−112=512 So, the probability that a randomly chosen friend takes biology or physics is 512.

The probability that a randomly chosen student plays in the band is 150. The probability that a randomly chosen student sings in the choir and plays in the band is 1200. What is the probability that a randomly chosen student sings in the choir, given that the student plays in the band?

If B is the event that a random student plays in the band, and C is the event that a random student sings in the choir. Then we are told P(B)=150 and P(C AND B)=1200. We are asked for P(C|B). Using the formula, we find that P(C|B)=P(C AND B)P(B)=1/2001/50=50200=14 So the probability that a randomly chosen student sings in the choir given that he or she plays in the band is 14.

Conditional probabilities may also deal with changes to the sample space. For example, if we are interested in the probability of drawing two diamonds in a row from a standard deck of playing cards, we have to consider that drawing the first card reduces the number of cards in the pile from 52 to 51. Problems will designate "with replacement" or "without replacement" to help you identify what is happening to the sample space as cards are being drawn.

If events (cards being drawn from a deck, marbles being chosen from a bag) are happening with replacement, an item from the population is chosen and then replaced before a second item is chosen. Here, the sample size does not change. If events are happening without replacement, the first item from the population that is chosen is not replaced before a second item is chosen. Here, the sample size changes.

In this section we will talk about how to calculate the conditional probability of events. This means that we will calculate the probability of an event happening given that another event has happened. For example, if we are asked to find the probability that a person misses their connecting flight given that their original flight was late, we are finding a conditional probability. We are finding the probability of one event, missing a connecting flight, based on the probability of another event already happening. Here, that the original flight was late is that other event. In this section we talk about the rule for finding probability P(A|B) in terms of the probabilities P(B) and P(A AND B). The idea is that for A to occur given that B has occurred, we need to know the probabilities of B happening as well as the probabilities of A and B happening. . P(A|B)=P(A AND B)P(B) This may be rewritten to help us find P(A AND B): P(A AND B)=P(A|B)P(B) The idea is that for both A and B to occur, we need B to occur, and we also need A to occur.

In this section we will talk about how to calculate the conditional probability of events. This means that we will calculate the probability of an event happening given that another event has happened. For example, if we are asked to find the probability that a person misses their connecting flight given that their original flight was late, we are finding a conditional probability. We are finding the probability of one event, missing a connecting flight, based on the probability of another event already happening. Here, that the original flight was late is that other event. In this section we talk about the rule for finding probability P(A|B) in terms of the probabilities P(B) and P(A AND B). The idea is that for A to occur given that B has occurred, we need to know the probabilities of B happening as well as the probabilities of A and B happening. . P(A|B)=P(A AND B)P(B) This may be rewritten to help us find P(A AND B): P(A AND B)=P(A|B)P(B) The idea is that for both A and B to occur, we need B to occur, and we also need A to occur.

A high school club is selling 15,000 raffle tickets at $10 each for a fundraiser. The table below shows the items given away, along with the value of the item and how many are available. Based on this information, what is the expected value of this raffle?

Recall, expected value can be found by multiplying the value by the probability for the event. Here, remember that you paid $10 for a raffle ticket. Therefore, you must subtract 10 from the value before finding the expected value. E(x)=990⋅1/15,000 + 490⋅2/15,000 + 290⋅2/15,000 +140⋅10/15,000 − 10⋅14,985/15,000 E(x)=0.066+0.0653+0.0387+0.0933−9.99 E(x)=−9.7267=−9.73 You can expect to lose $9.73 for each raffle ticket.

Let R be the event that a randomly chosen person has visited Rome, Italy. Let G be the event that a randomly chosen person has visited Greece. Place the correct event in each response box below to show: Given that the person has visited Rome, Italy, the probability that a randomly chosen person has visited Greece.

Remember that in general, P(A|B) is read as "The probability of A given B". Here we want to know the probability that a person has visited Greece given that the person has visited Rome, Italy, so the correct answer is P(G|R).

Question The probability that a randomly chosen college student smokes is 0.2. The probability that a randomly chosen college student is an athlete is 0.15. The probability that a randomly chosen college student smokes given that the student is an athlete is 0.1. What is the probability that a randomly chosen college student is a smoker and an athlete?

Solution Let S be the event that a randomly chosen student smokes. Let A be the event that a randomly chosen student is an athlete. We are told P(S)=0.2, P(A)=0.15, and P(S|A)=0.1. We are asked to find P(S AND A). Using what we are told and the formula for the compound probability above, we find that P(S AND A)=P(S|A)P(A)=(0.1)⋅(0.15)=0.015 So the probability that a randomly chosen student is both an athlete and a smoker is0.015. Note that we did not use the information about P(S). Sometimes you may be given extraneous information in problems, so you need to be careful to apply the formula correctly.

If A and B are independent events with P(A)=0.60 and P(A AND B)=0.30, find P(B). Give your answer as a percent, rounded to two decimal places if necessary.

Correct answers:$50\%$50%​ Remember that for independent events, P(A AND B)=P(A)P(B) Solving for P(B), we find thatP(B)=P(A AND B)P(A)Now, plugging in the values we are given, we find thatP(B)=0.300.60=0.50 To express this answer as a percent, we multiply by 100: 0.50×100=50%

If A and B are events with P(A)=0.8, P(A OR B)=0.87, P(A AND B)=0.23, find P(B).

$0.3$0.3​ First, it is helpful to write down the addition rule for probabilities: P(A OR B)=P(A)+P(B)−P(A AND B) Now, rearranging this, we find thatP(B)=P(A OR B)+P(A AND B)−P(A)Plugging in the known values, we findP(B)=0.87+0.23−0.8=0.3

A poker game requires players to draw a card and roll a die. The combination of card and die will determine if a player wins. There are: 13 different types of cards: Ace, 2,3,4,5,6,7,8,9,10, Jack, Queen, King 6 different numbers, one on each face of a die: 1,2,3,4,5,6 How many different combinations are there of one card and one die roll?

Correct answers:$78$78​ The Fundamental Counting Principle tells us that we multiple the number of cards by the number of faces on the die to get the total number of possibilities: Possibilities=13×6=78

In a game show, contestants are given the opportunity to win a new car if they correctly choose the winning door three times in a row. In the first round, they must choose between 3 doors, one of which is labeled "win." In the second round and third rounds, they will choose from 2 doors, with one labeled "win." Their options are shown in the tree diagram below. They are blind folded, and are not given an opportunity to see the doors beforehand. What is the probability that a contestant will NOT win a new car?

Correct answer: 11/12 To win the new car, a contestant must choose a "win" door three times in a row. We can see from the tree diagram that there is only one possibility of doing this, but there are 12 possible outcomes. So, there are 12−1=11 ways to NOT win. The probability of a contestant not winning is then 11/12.

Two friends are both pregnant, and find out they are each expecting twins! Let A be the event that one friend is pregnant with identical twins, and note that P(A)=0.0045. Let B be the event that the other friend is pregnant with fraternal twins, and note that P(B)=0.01. A and B are independent events. What is the probability that one friend is pregnant with identical twins, and one friend is pregnant with fraternal twins? Give your answer as a percent, rounded to four decimal places if necessary.

Correct answers:$0.0045\%$0.0045%​ Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=0.0045⋅0.01=0.000045 To express our answer as a percent, we multiply by 100: 0.000045×100=0.0045%

The probability that a student will take loans to pay for their undergraduate education is 0.85, and the probability that a student will go to graduate school given that the student took loans to pay for their undergraduate education is 0.13. What is the probability that a student will go to graduate school and take loans to pay for their undergraduate education? Round your answer to three decimal places.

Correct answers:$0.111$0.111​ Remember the multiplication rule for conditional probability: P(B AND A)=P(B|A)P(A) So if we think of A as the event of a student taking loans to pay for their undergraduate education and B as being the event of a student going to graduate school, then we can plug in the known information to find P(B AND A)=(0.13)(0.85)≈0.111

An electric company has to track on-time payments and late payments for each customer monthly. It is impossible for a customer to pay on-time and late each month. If the probability that a customer pays on-time each month is 0.55, and the probability that a customer pays late or on-time each month is 0.82, what is the probability that a customer pays late each month?

Correct answers:$0.27$0.27​ Because it is impossible for a customer to pay on-time and late each month, we see that they are mutually exclusive events. Therefore, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) Rearranging, we find thatP(A)=P(A OR B)−P(B)So plugging in the given values, we findP(A)=0.82−0.55=0.27 The probability that a customer pays late each month is 0.27.

A hospital conducts a study of men and women to see if they are satisfied with their current weight. Let A be the event that a randomly selected woman is satisfied with her weight and B be the event that a randomly selected man is satisfied with his weight. The probabilities of these events are as follows: P(A)=56% and P(B)=85% A and B are independent events. What is P(A AND B), the probability that a randomly selected man and woman will both be satisfied with their weight? Give your answer as a decimal, rounded to three decimal places.

Correct answers:$0.476$0.476​ Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=56%⋅85%=0.56⋅0.85=0.476

If a police officer pulls over someone for speeding, the police officer can either give a ticket or a warning, so it is impossible for a police officer to give a ticket and a warning for speeding. If the probability that a police officer will give a warning for speeding is 0.03, and the probability that a police officer will give a ticket or a warning for speeding is 0.52, what is the probability that a police officer will give a ticket for speeding?

Correct answers:$0.49$0.49​ Because it is impossible for a police officer to give a ticket and a warning for speeding, we see that they are mutually exclusive events. Therefore, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) Rearranging, we find thatP(A)=P(A OR B)−P(B)So plugging in the given values, we findP(A)=0.52−0.03=0.49 The probability that a police officer will give a ticket for speeding is 0.49.

Suppose A and B are mutually exclusive events, and that P(B)=0.03 and P(A OR B)=0.52. Find P(A).

Correct answers:$0.49$0.49​ Remember that when A and B are mutually exclusive events, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) Rearranging, we find thatP(A)=P(A OR B)−P(B)So plugging in the given values, we findP(A)=0.52−0.03=0.49

Randy wants to either ride share to work or drive his own car to work, but it is impossible for Randy to ride share and drive his own car in one trip. If the probability that Randy ride shares is 0.22, and the probability that Randy drives his own car is 0.42, what is the probability that Randy ride shares or drives his own car to work?

Correct answers:$0.64$0.64​ Because it is impossible for Randy to ride share and drive his own car in one trip, we see that they are mutually exclusive events. Therefore, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) So we find thatP(A OR B)=P(A)+P(B)=0.22+0.42=0.64 The probability that Randy ride shares or drives his own car to work is 0.64.

Given that the probability of a student taking a statistics class is 0.83, and the probability of a student taking a calculus class and a statistics class is 0.66, what is the probability of a student taking a calculus class given that the student takes a statistics class? Round your answer to three decimal places.

Correct answers:$0.795$0.795​ Remember the multiplication rule for conditional probability: P(A AND B)=P(A|B)P(B) Rearranging this, we findP(A|B)=P(A AND B)P(B)So if we think of A as being the event a student taking a calculus class and B as being the event a student taking a statistics class, then we can plug in the known information to findP(A|B)=0.660.83≈0.795

There are two known issues with a certain model of new car. The first issue, A, occurs with a probability of P(A)=0.1. B is another known issue with the car. If it is known that either event occurs with a probability of P(A OR B)=0.93, and that both events occur with a probability of P(A AND B)=0.07, calculate P(B).

Correct answers:$0.9$0.9​ Begin with the Addition Rule: P(A OR B)=P(A)+P(B)−P(A AND B) Rearranging to solve for P(B), we find that P(B)=P(A OR B)−P(A)+P(A AND B)=0.93−0.1+0.07=0.9 So P(B)=0.9.

A student is mapping out their options for their post-secondary education. They list out all of their considerations. How many possible combinations of college, master's degree, and further education are there?

Correct answers:$12$12​ The Fundamental Counting Principle tells us that we multiple the number of choices in each section by one another to get the total number of possibilities. Possibilities=2×2×3=12

Two companies offer their employees retirement plans. Let A be the event that a randomly chosen employee from the first company chooses to invest in the retirement plan. Let B be the event that a randomly chosen employee from the second company chooses to invest in the retirement plan. The probabilities of them investing in these plans are as follows: P(A)=0.30 and P(B)=0.48 You choose one employee randomly from each company. If A and B are independent events, what is P(A AND B), the probability that they both invest in the retirement plan? Give your answer as a percent, rounded to two decimal places if necessary.

Correct answers:$14.4\%$14.4%​ Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=0.30⋅0.48=0.144 To express our answer as a percent, we multiply by 100: 0.144×100=14.4%

Complete the following contingency table and find the number of students who both do not play sports AND do not play an instrument.

Correct answers:$18\text{ students}$18 students​ By using the known totals along the rows and columns you can fill in the rest of the contingency table. For example, looking at the first row in the table, we know that 19 added to the unknown number in the middle is 47, so that unknown number is 28. Continuing in this way, we can fill in the entire table: StudentsPlay an instrumentDo not play an instrumentTotalPlay sports192847Do not play sports111829Total304676 From this, we can see that the number of students who both do not play sports and do not play an instrument is 18.

A probability distribution for a random variable X is given by P(X=x)=cx, for x=1,2,3,4 and c is a constant. Find E(X), the expected value of X.

Correct answers:$3$3​ The first thing we must do is calculate c. This can be done by noting that all of the probabilities must add up to 1. Therefore c+2c+3c+4c=1, which solves to give c=0.1. Using this, we can now say thatP(X=1)=0.1,P(X=2)=0.2,P(X=3)=0.3,P(X=4)=0.4.We can now compute E(X), thereby givingE(X)=1⋅0.1+2⋅0.2+3⋅0.3+4⋅0.4=3.

Fill in the following contingency table and find the number of students who both play sports AND play an instrument.

Correct answers:$30$30​ By using the known totals along the rows and columns you can fill in the rest of the contingency table. For example, looking at the second row in the table, we know that 23 added to the unknown number in the middle is 48, so that unknown number is 25. Continuing in this way, we can fill in the entire table: StudentsplaysportsdonotplaysportsTotalplayaninstrument302353donotplayaninstrument212546Total514899 From this, we can see that the number of students who both play sports and play an instrument is 30.

Given that P(B AND A)=0.07 and P(B|A)=0.20, what is P(A)? Express your answer as a percent, rounded to two decimal places if necessary.

Correct answers:$35\%$35%​ Remember the multiplication rule for conditional probability: P(B AND A)=P(B|A)P(A) Rearranging, we find thatP(A)=P(B AND A)P(B|A)Plugging in the values we were given, we find thatP(A)=0.07/0.20=0.35 To express our answer as a percent, we multiply by 100: 0.35×100=35%

Given that P(A AND B)=0.29 and P(A|B)=0.67, what is P(B)? Give your answer as a percent. Round to two decimal places.

Correct answers:$43.28\ \%$43.28 %​ Remember the multiplication rule for conditional probability: P(A AND B)=P(A|B)P(B) Rearranging, we find thatP(B)=P(A AND B)P(A|B)Plugging in the values we were given, we find that P(B)=0.29*0.67≈0.4328 To convert the decimal to a percent, we can multiply by 100: 0.4328×100=43.28%

The probability that a child watches cartoons after school is 0.56. The probability that the child requests a certain toy of a cartoon character given that the child watches cartoons after school is 0.89. What is the probability that a child will request a toy and watch cartoons after school? Give your answer as a percent, rounded to two decimal places if necessary.

Correct answers:$49.84\%$49.84%​ Remember the multiplication rule for conditional probability: P(A AND B)=P(A|B)P(B) So if we think of A as being the event of a child requesting a toy and B as being the event of a watching cartoons after school, then we can plug in the known information to findP(A AND B)=(0.89)(0.56)≈0.4984 To express our answer as a percent, we multiply by 100: 0.4984×100=49.84%

If you roll a fair die and then roll a second fair die, what is the probability that you roll a 1, 2, or 3 on the first die and roll a 5 on the second die? Write your answer as a fraction

Correct answers:$\frac{1}{12}$112​​ Note that these are independent events because the outcome when you roll a fair die does not affect the outcome when you roll a second fair die. So by the multiplication rule for independent events, we can take the probability of each event and multiply them.The probability that you roll a 1, 2, or 3 on the first die is 36, and the probability that you roll a 5 on the second die is 16. The probability that you roll a 1, 2, or 3 on the first die and roll a 5 on the second die is 36⋅16=112

Adam loves to visit his local coffee shop. When there, his two favorite drinks are cafe latte and espresso. Let A represent the event that he drinks a cafe latte and B represent the event that he drinks an espresso. If A and B are independent events with P(A)=0.5 and P(B)=0.8, find P(A AND B). Give your answer as a fraction in simplest form.

Correct answers:$\frac{2}{5}$25​​ Remember that for independent events, P(A AND B)=P(A)P(B) So, plugging in the values we are given, we find thatP(A AND B)=(0.50)(0.80)=0.4 To express this answer as a fraction in simplest form, we have 0.4=410=25 The probability Adam goes to the coffee shop and drinks both a cafe latte and an espresso is 25.

The event that it is a sunny day and the event that you are given a speeding ticket are independent events. Let A be the event that it is a sunny day with P(A)=35. Let B be the event that you are given a speeding ticket with P(B)=120. What is the probability that it is a sunny day and you are given a speeding ticket? Give your answer as a fraction in simplest form.

Correct answers:$\frac{3}{100}$3100​​ Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=35⋅120=3/100

First year nursing students are randomly assigned to classes in a nursing school. Let A be the event that they are assigned to an Anatomy course and B be the event that they are assigned to a Psychology course. The probabilities of them being assigned to these courses are as follows: P(A)=23 and P(B)=25 If A and B are independent events, what is P(A AND B), the probability that a student will be assigned to both an Anatomy and a Psychology course? Give your answer as a fraction.

Correct answers:$\frac{4}{15}$415​​ Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=2/3⋅2/5=4/15

The probability that a car has a certain factory defect is 8/25. The probability that a car has a certain factory defect and needs an oil change is 7/50. What is the probability that a car needs an oil change given that it has a certain factory defect? Give your answer as a fraction in simplest form.

Correct answers:$\frac{7}{16}$716​​ Remember the multiplication rule for conditional probability: P(A AND B)=P(A|B)P(B) Rearranging, we see thatP(A|B)=P(A AND B)P(B) So if we think of A as being the event that a car has a factory defect and B as being the event that the car needs an oil change, then we can plug in the known information to find P(A|B)=750825=7/16

Let A be the event that a person who lives in New York City is married, and note that P(A)=25. Let B be the event that a person who lives in Honolulu is married, and note that P(B)=720. Events A and B are independent. What is the probability that a randomly chosen person from New York City and a random chosen person from Honolulu will be married? Give your answer as a fraction in simplest form.

Correct answers:$\frac{7}{50}$750​​ Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=2/5⋅7/20=7/50

"Marble Madness" is a local carnival game, costing $2. There are 100 total marbles in a bag: 2 red, 8 orange, 10 yellow, 30 green, 30 blue, and 20 black. If a red marble is pulled, you win $6, an orange marble wins $4, and a yellow marble wins $2. A green, blue, and black marble result in a loss of the $2 cost to play. What is the expected value of a marble pull?

Correct answers:1$-1.36$−1.36​ Recall, expected value is found by multiplying the value times the probability of the event. E(x)=4⋅2100+2⋅8100+0⋅10100−2⋅80100 E(x)=0.08+0.16+0−1.6 E(x)=−1.36 You can expect to lose $1.36 each time you pull a marble.

A bicycle shop sells 10,000 raffle tickets at $5 each, raising money for a community park. The table below describes the prizes, value of each prize, and how many are being given away. PrizeValueNumber to be Given AwayBicycle and Helmet$18020Bicycle$15015Helmet$3010 What is the expected value of the raffle? Round to the nearest cent. Do not round until the final calculation.

Correct answers:1$-4.39$−4.39​ Recall, expected value is found by multiplying the value by the probability of the event. Remember, in this example, you paid $5 for the raffle ticket, so you must subtract 5 from the 'value' in order to find the expected value. E(x)=175⋅20/10,000 + 145⋅15/10,000 + 25⋅10/10,000 − 5⋅9,955/10,000 E(x)=0.35+0.2175+0.025−4.9775 E(x)=−4.385−−4.39 You can expect to lose $4.39 for each raffle ticket purchased.

Janelle is playing craps and places a $5 bet on the number 8 with a probability of winning at 536. However, the probability of losing is 16 and happens if a 7 is rolled. Any other number rolled means she doesn't win or lose and the game continues. If an 8 is rolled, she will win $6 (and keep her initial bet) but if a 7 is rolled, she loses all $5. What is the expected value for Janelle on a roll in this game of craps? Round to the nearest cent. Do not round until the final calculation.

Correct answers:1$0$0​ Recall, the expected value can be calculated by multiplying the value by the probability of that event. Here, Janelle can win $6 or lose $5. E(x)=6⋅536−5⋅636 E(x)=30/36−30/36 E(x)=0 Therefore, Janelle can expect to break even on each roll during a game of craps, if she only bets on the number 8.

Janet plays a dice game. If she rolls a 1, she wins $3 while if she rolls a 2, she wins $1. If she rolls a 3,4,or5, she loses $1 and if a 6 is rolled, she doesn't win or lose. Find the expected value of this dice game. Round to the nearest cent. Do not round until the final calculation.

Correct answers:1$0.17$0.17​ Recall, expected value is found by multiplying the value by the probability of the event. Here Janet can win 3, win 1, lose 1, or break even. E(x)=3⋅1/6+1⋅1/6−1⋅1/2+0⋅1/6 E(x)=3/6+1/6−3/6+0=16 E(x)=16=0.166666...=0.17 Janet can expect to win $0.17 each time she rolls the dice.

A university is randomly assigning its 150 first-year students to writing classes or study skills seminars. 50 of the students will be randomly selected to take the writing seminar the first semester, and 15 of the students will be randomly assigned to the study skills seminar for the second semester. Only 5 students will be allowed to take both seminars. What is the probability that a student will be assigned to the study skills seminar given that they were chosen to participate in the writing seminar?

Let A be the event that the student is randomly selected to take the writing seminar, and note that P(A)=50150=13. Let B be the event that the student is selected to take the study skills seminar, and note that P(B)=15150=110. P(A AND B), the probability that a student takes both the writing and the study skills seminars, is 5150=130. Now to calculate P(B|A), the probability that a student will be assigned to the study skills seminar given that they were chosen to participate in the writing seminar: P(B|A)=P(B AND A)P(A)=13013=330=110 Conditional probabilities may also deal with changes to the sample space. For example, if we are interested in the probability of drawing two diamonds in a row from a standard deck of playing cards, we have to consider that drawing the first card reduces the number of cards in the pile from 52 to 51. Problems will designate "with replacement" or "without replacement" to help you identify what is happening to the sample space as cards are being drawn. If events (cards being drawn from a deck, marbles being chosen from a bag) are happening with replacement, an item from the population is chosen and then replaced before a second item is chosen. Here, the sample size does not change. If events are happening without replacement, the first item from the population that is chosen is not replaced before a second item is chosen. Here, the sample size changes.

A bag of candy contains 3 blue candies, 4 green candies, and 6 red candies. If you draw two candies, one at a time and without replacement, what is the probability that you will draw a green candy and a blue candy?

Let G be the event of drawing a green candy on the first draw and B be the event of drawing a blue candy on the second draw. Then P(B|G) is the probability of drawing a blue candy given that the first candy was green, and P(B AND G) is the probability of drawing a green and a blue candy. P(G)=413; P(B|G)=312=14. Notice that here there are 3 possible blue candies and 12 candies left after drawing the first green candy (since we drew without replacement). We want to know P(B AND G), the probability of drawing a green and a blue candy. The Multiplication Rule for Conditional Probabilities gives that: P(B|G)=P(B AND G)P(G) Rearranging using algebra, we have: P(B AND G)=P(B|G)⋅P(G) Plugging in the probabilities from above, we have P(B AND G)=14⋅413=113

Employees at a company are randomly assigned certain shifts during the busy holiday season. Let A represent the shift between the hours of 8 a.m. and 12 p.m., and B represent the shift between the hours of 12 and 4 p.m. If the shifts are assigned with the following probabilities: P(A)=0.28; P(B)=0.83; P(A OR B)=0.93 What is P(A AND B), the probability that an employee will randomly be assigned both shifts?

Solution: Here, we are given slightly different information than in the rule above, but note that we can rearrange the rule to solve for P(A AND B): P(A AND B)=P(A)+P(B)−P(A OR B)=0.28+0.83−0.93=0.18 So we find that P(A AND B)=0.18. This means that the probability that an employee will randomly be assigned both shifts is 0.18, or 18%.

Question: A university is randomly assigning its 150 first-year students to writing classes or study skills seminars. 50 of the students will be randomly selected to take the writing seminar the first semester, and 15 of the students will be randomly assigned to the study skills seminar for the second semester. Only 5 students will be allowed to take both seminars. What is the probability that a student will be assigned to the study skills seminar given that they were chosen to participate in the writing seminar?

Solution: Let A be the event that the student is randomly selected to take the writing seminar, and note that P(A)=50150=13. Let B be the event that the student is selected to take the study skills seminar, and note that P(B)=15150=110. P(A AND B), the probability that a student takes both the writing and the study skills seminars, is 5150=130. Now to calculate P(B|A), the probability that a student will be assigned to the study skills seminar given that they were chosen to participate in the writing seminar: P(B|A)=P(B AND A)P(A)=1/30//1/3 =3/30=1/10

A game requires that players draw a blue card and a red card to determine the number of spaces they can move on a turn. Let A represent drawing a red card, and with four possibilities: 1,2,3, and 4. Let B represent drawing a blue card, and notice that there are three possibilities 1,2, and 3. A and B are independent events. What is the probability that a player will draw a red 2 and a blue 3?

Solution: The probability of a player drawing a red 2 is 14, and the probability of a player drawing a blue 3 is 13. We know that the events are independent, so P(A AND B)=P(A)⋅P(B)=1/4⋅1/3=1/12