Ch 3 Quizlet: Conditional Probability

3.64 Independent flips of a coin that lands on heads with probability p are made. What is the probability that the first four outcomes are (b) T, H, H, H?

(1-p)p^3

3.13 A recent college graduate is planning to take the first three actuarial examinations in the coming summer. She will take the first actuarial exam in June. If she passes that exam, then she will take the second exam in July, and if she also passes that one, then she will take the third exam in September. If she fails an exam, then she is not allowed to take any others. The probability that she passes the first exam is .9. If she passes the first exam, then the conditional probability that she passes the second one is .8, and if she passes both the first and the second exams, then the conditional probability that she passes the third exam is .7. (b) Given that she did not pass all three exams, what is the conditional probability that she failed the second exam? (probability she passes all 3 exams is .504)

.03629

3.22 A total of 500 married working couples were polled about their annual salaries, with the following information resulting: Husband Wife Less than $25,000 More than$25,000 Less than $25,000 212 198 More than $25,000 36 54 For instance, in 36 of the couples, the wife earned more and the husband earned less than $25,000. If one of the couples is randomly chosen, what is (c) the conditional probability that the wife earns more than $25,000 given that the husband earns less than this amount?

.1452

3.22 A total of 500 married working couples were polled about their annual salaries, with the following information resulting: Husband Wife Less than $25,000 More than$25,000 Less than $25,000 212 198 More than $25,000 36 54 For instance, in 36 of the couples, the wife earned more and the husband earned less than $25,000. If one of the couples is randomly chosen, what is (b) the conditional probability that the wife earns more than $25,000 given that the husband earns more than this amount?

.2143

3.20 A total of 48 percent of the women and 37 percent of the men who took a certain "quit smoking" class remained nonsmokers for at least one year after completing the class. These people then attended a success party at the end of a year. If 62 percent of the original class was male, (a) what percentage of those attending the party were women? (b) what percentage of the original class attended the party?

.443

3.22 A total of 500 married working couples were polled about their annual salaries, with the following information resulting: Husband Wife Less than $25,000 More than$25,000 Less than $25,000 212 198 More than $25,000 36 54 For instance, in 36 of the couples, the wife earned more and the husband earned less than $25,000. If one of the couples is randomly chosen, what is (a) the probability that the husband earns less than $25,000?

.496

3.13 A recent college graduate is planning to take the first three actuarial examinations in the coming summer. She will take the first actuarial exam in June. If she passes that exam, then she will take the second exam in July, and if she also passes that one, then she will take the third exam in September. If she fails an exam, then she is not allowed to take any others. The probability that she passes the first exam is .9. If she passes the first exam, then the conditional probability that she passes the second one is .8, and if she passes both the first and the second exams, then the conditional probability that she passes the third exam is .7. (a) What is the probability that she passes all three exams?

.504

3.17 Ninety-eight percent of all babies survive delivery. However, 15 percent of all births involve Cesarean (C) sections, and when a C section is performed, the baby sur- vives 96 percent of the time. If a randomly chosen preg- nant woman does not have a C section, what is the proba- bility that her baby survives?

.9835

3.8 A couple has 2 children. What is the probability that both are girls if the older of the two is a girl?

1/2

3.10 Three cards are randomly selected, without replace- ment, from an ordinary deck of 52 playing cards. Compute the conditional probability that the first card selected is a spade given that the second and third cards are spades.

11/50

3.7 The king comes from a family of 2 children. What is the probability that the other child is his sister?

2/3

5.58 A parallel system functions whenever at least one of its components works. Consider a parallel system of n components, and suppose that each component works independently with probability 1/2 . Find the conditional probability that component 1 works given that the system is functioning.

2^(n-1) / 2^n - 1

3.5 An urn contains 6 white and 9 black balls. If 4 balls are to be randomly selected without replacement, what is the probability that the first 2 selected are white and the last 2 black?

6/91 P(AB)= P(B I A)P(A) Let A=the first two selected balls are white B=the second two selected balls are black

3.9 Consider 3 urns. Urn A contains 2 white and 4 red balls, urn B contains 8 white and 4 red balls, and urn C con- tains 1 white and 3 red balls. If 1 ball is selected from each urn, what is the probability that the ball chosen from urn A was white given that exactly 2 white balls were selected?

7/11

3.16 An ectopic pregnancy is twice as likely to develop when the pregnant woman is a smoker as it is when she is a nonsmoker. If 32 percent of women of childbearing age are smokers, what percentage of women having ectopic pregnancies are smokers?

= .485

3.65 The color of a person's eyes is determined by a single pair of genes. If they are both blue-eyed genes, then the person will have blue eyes; if they are both brown-eyed genes, then the person will have brown eyes; and if one of them is a blue-eyed gene and the other a brown-eyed gene, then the person will have brown eyes. (Because of the latter fact, we say that the brown-eyed gene is dominant over the blue-eyed one.) A newborn child independently receives one eye gene from each of its parents, and the gene it receives from a parent is equally likely to be either of the two eye genes of that parent. Suppose that Smith and both of his parents have brown eyes, but Smith's sister has blue eyes. (b) Suppose that Smith's wife has blue eyes. What is the probability that their first child will have blue eyes?

= 1/3

3.50 Each of 2 cabinets identical in appearance has 2 drawers. Cabinet A contains a silver coin in each drawer, and cabinet B contains a silver coin in one of its draw- ers and a gold coin in the other. A cabinet is randomly selected, one of its drawers is opened, and a silver coin is found. What is the probability that there is a silver coin in the other drawer?

= 2/3

3.43 A deck of cards is shuffled and then divided into two halves of 26 cards each. A card is drawn from one of the halves; it turns out to be an ace. The ace is then placed in the second half-deck. The half is then shuffled, and a card is drawn from it. Compute the probability that this drawn card is an ace. Hint: Condition on whether or not the interchanged card is selected.

= 5/54

3.64 Independent flips of a coin that lands on heads with probability p are made. What is the probability that the first four outcomes are (a) H, H, H, H?

= p^4

3.44 Twelve percent of all U.S. households are in California. A total of 1.3 percent of all U.S. households earn more than $250,000 per year, while a total of 3.3 per- cent of all California households earn more than $250,000 per year. (a) What proportion of all non-California households earn more than $250,000 per year?

=.0103

3.31 There are 15 tennis balls in a box, of which 9 have not previously been used. Three of the balls are randomly chosen, played with, and then returned to the box. Later, another 3 balls are randomly chosen from the box. Find the probability that none of these balls has ever been used.

=.0893

3.44 Twelve percent of all U.S. households are in California. A total of 1.3 percent of all U.S. households earn more than $250,000 per year, while a total of 3.3 per- cent of all California households earn more than $250,000 per year. (b) Given that a randomly chosen U.S. household earns more than $250,000 per year, what is the probability it is a California household?

=.305

3.37 With probability .6, the present was hidden by mom; with probability .4, it was hidden by dad. When mom hides the present, she hides it upstairs 70 percent of the time and downstairs 30 percent of the time. Dad is equally likely to hide it upstairs or downstairs. (b) Given that it is downstairs, what is the probability it was hidden by dad?

=.5263

3.37 With probability .6, the present was hidden by mom; with probability .4, it was hidden by dad. When mom hides the present, she hides it upstairs 70 percent of the time and downstairs 30 percent of the time. Dad is equally likely to hide it upstairs or downstairs. (a) What is the probability that the present is upstairs?

=.62

3.35 On rainy days, Joe is late to work with probability .3; on nonrainy days, he is late with probability .1. With probability .7, it will rain tomorrow. (b) Given that Joe was early, what is the conditional prob- ability that it rained?

=.64

3.51 Prostate cancer is the most common type of cancer found in males. As an indicator of whether a male has prostate cancer, doctors often perform a test that measures the level of the prostate-specific antigen (PSA) that is pro- duced only by the prostate gland. Although PSA levels are indicative of cancer, the test is notoriously unreliable. Indeed, the probability that a noncancerous man will have an elevated PSA level is approximately .135, increasing to approximately .268 if the man does have cancer. If, on the basis of other factors, a physician is 70 percent certain that a male has prostate cancer, what is the conditional proba- bility that he has the cancer given that b) the test did not indicate an elevated PSA level?

=.6638

3.52 Suppose that an insurance company classifies people into one of three classes: good risks, average risks, and bad risks. The company's records indicate that the probabilities that good-, average-, and bad-risk persons will be involved in an accident over a 1-year span are, respectively, .05, .15, and .30. If 20 percent of the population is a good risk, 50 percent an average risk, and 30 percent a bad risk, what proportion of people have accidents in a fixed year? If policyholder A had no accidents in 2012, what is the prob- ability that he or she is a good risk? is an average risk?

=.7455

3.35 On rainy days, Joe is late to work with probability .3; on nonrainy days, he is late with probability .1. With probability .7, it will rain tomorrow. (a) Find the probability that Joe is early tomorrow.

=.76

3.51 Prostate cancer is the most common type of cancer found in males. As an indicator of whether a male has prostate cancer, doctors often perform a test that measures the level of the prostate-specific antigen (PSA) that is pro- duced only by the prostate gland. Although PSA levels are indicative of cancer, the test is notoriously unreliable. Indeed, the probability that a noncancerous man will have an elevated PSA level is approximately .135, increasing to approximately .268 if the man does have cancer. If, on the basis of other factors, a physician is 70 percent certain that a male has prostate cancer, what is the conditional proba- bility that he has the cancer given that (a) the test indicated an elevated PSA level?

=.8224

3.38 Stores A, B, and C have 50, 75, and 100 employees, respectively, and 50, 60, and 70 percent of them respec- tively are women. Resignations are equally likely among all employees, regardless of sex. One woman employee resigns. What is the probability that she works in store C?

=0.5

3.1 Two fair dice are rolled. What is the conditional prob- ability that at least one lands on 6 given that the dice land on different numbers?

=1/3

3.3 Use Equation (2.1) to compute in a hand of bridge the conditional probability that East has 3 spades given that North and South have a combined total of 8 spades. (52 cards, each player gets 13 cards)

We need: P(A)= IAI/ISI where ∣X∣ denotes the number of elements in X The number of elements in S here is ∣S ∣=(52 choose 13​)(39 choose 13​)(26 ​choose 13)=52!​ / (13!)^4 (when choosing cards for North then for the East then for the South, the remaining cards go to West) We also need the formula for conditional probability: P(B IA)= P(AB)/A Probability= .339

3.62 A simplified model for the movement of the price of a stock supposes that on each day the stock's price either moves up 1 unit with probability p or moves down 1 unit with probability 1 − p. The changes on different days are assumed to be independent. (a) What is the probability that after 2 days the stock will be at its original price? (b) What is the probability that after 3 days the stock's price will have increased by 1 unit? (c) Given that after 3 days the stock's price has increased by 1 unit, what is the probability that it went up on the first day?

a) 2p(1-p) b) 3p^2(1-p) c) 2/3

3.42 Consider a sample of size 3 drawn in the following manner: We start with an urn containing 5 white and 7 red balls. At each stage, a ball is drawn and its color is noted. The ball is then returned to the urn, along with an addi- tional ball of the same color. Find the probability that the sample will contain exactly (a) 0 white balls; (b) 1 white ball; (c) 3 white balls; (d) 2 white balls.

a) 3/13 b) 5/39 c) 5/52 d) 85/156

3.24 Urn I contains 2 white and 4 red balls, whereas urn II contains 1 white and 1 red ball. A ball is randomly chosen from urn I and put into urn II, and a ball is then randomly selected from urn II. What is (a) the probability that the ball selected from urn II is white? (b) the conditional probability that the transferred ball was white given that a white ball is selected from urn II?

a) 4/9 b) 1/2

3.23 A red die, a blue die, and a yellow die (all six sided) are rolled. We are interested in the probability that the number appearing on the blue die is less than that appear- ing on the yellow die, which is less than that appearing on the red die. That is, with B, Y, and R denoting, respec- tively, the number appearing on the blue, yellow, and red die, we are interested in P(B < Y < R). (a) What is the probability that no two of the dice land on the same number? (b) Given that no two of the dice land on the same num- ber, what is the conditional probability that B < Y < R? (c) What is P(B < Y < R)?

a) 5/9 b) 1/6 c) 5/54

3.21 Fifty-two percent of the students at a certain college are females. Five percent of the students in this college are majoring in computer science. Two percent of the students are women majoring in computer science. If a student is selected at random, find the conditional probability that (a) the student is female given that the student is majoring in computer science; (b) this student is majoring in computer science given that the student is female.

a) = .4 b) = .03846

3.63 Suppose that we want to generate the outcome of the flip of a fair coin, but that all we have at our disposal is a biased coin that lands on heads with some unknown proba- bility p that need not be equal to 1/2 . Consider the following procedure for accomplishing our task: 1. Flip the coin. 2. Flip the coin again. 3. If both flips land on heads or both land on tails, return to step 1. 4. Let the result of the last flip be the result of the experi- ment. (a) Show that the result is equally likely to be either heads or tails. (b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different and then lets the result be the outcome of the final flip?

a) =1/2 b) This method takes the result of the first coin toss that has not been seen yet. So if the tosses are heads, heads, heads, tails, the final result is tails. That means tails is the result if and only if the first throw is heads, then the first different result is tails. Now the result has distribution (P(tails))=p, and (P(heads))= 1−p. Reverse of the distribution of the coin.

5.59 If you had to construct a mathematical model for events E and F, as described in parts (a) through (e), would you assume that they were independent events? Explain your reasoning. (a) E is the event that a businesswoman has blue eyes, and F is the event that her secretary has blue eyes. (b) E is the event that a professor owns a car, and F is the event that he is listed in the telephone book. (c) E is the event that a man is under 6 feet tall, and F is the event that he weighs more than 200 pounds. (d) E is the event that a woman lives in the United States, and F is the event that she lives in the Western Hemi- sphere. (e) E is the event that it will rain tomorrow, and F is the event that it will rain the day after tomorrow.

a) independent b) independent c) dependent d) dependent e) dependent

3.67 Barbara and Dianne go target shooting. Suppose that each of Barbara's shots hits a wooden duck target with probability p1, while each shot of Dianne's hits it with probability p2. Suppose that they shoot simultaneously at the same target. If the wooden duck is knocked over (indi- cating that it was hit), what is the probability that (a) both shots hit thterm-83e duck? (b) Barbara's shot hit the duck? What independence assumptions have you made?

a) p1​p2​​ / (p1 + p2 - p1p2) b) p1 / (p1 + p2 - p1p2) Independence: Both shoot at the duck at the same time, and their hits are independent.

3.15 An urn initially contains 5 white and 7 black balls. Each time a ball is selected, its color is noted and it is replaced in the urn along with 2 other balls of the same color. Compute the probability that (a) the first 2 balls selected are black and the next 2 are white; (b) of the first 4 balls selected, exactly 2 are black.

a. .0456 b. .2734

3.47 There is a 30% chance that A can fix her busted computer. If A cannot, then there is a 40% chance that friend B can fix it. a) Find the probability it will be fixed by either A or B b) If it is fixed, what is the probability it will be fixed by B?

a= .58 b= .69

3.68 A and B are involved in a duel. The rules of the duel are that they are to pick up their guns and shoot at each other simultaneously. If one or both are hit, then the duel is over. If both shots miss, then they repeat the process. Suppose that the results of the shots are independent and that each shot of A will hit B with probability pA, and each shot of B will hit A with probability pB. What is (a) the probability that A is not hit?

pa / (pa + pb - papb)

3.68 A and B are involved in a duel. The rules of the duel are that they are to pick up their guns and shoot at each other simultaneously. If one or both are hit, then the duel is over. If both shots miss, then they repeat the process. Suppose that the results of the shots are independent and that each shot of A will hit B with probability pA, and each shot of B will hit A with probability pB. What is (b) the probability that both duelists are hit?

papb​​ / (pa + pb - papb)

3.27 The following method was proposed to estimate the number of people over the age of 50 who reside in a town of known population 100,000: "As you walk along the streets, keep a running count of the percentage of peo- ple you encounter who are over 50. Do this for a few days; then multiply the percentage you obtain by 100,000 to obtain the estimate." Comment on this method. Hint: Let p denote the proportion of people in the town who are over 50. Furthermore, let α1 denote the propor- tion of time that a person under the age of 50 spends in the streets, and let α2 be the corresponding value for those over 50. What quantity does the method suggested esti- mate? When is the estimate approximately equal to p?

when a1=a2 this estimate is approx equal to p

What do the odds of Event A tell us? How do we calculate?

That is, the odds of an event A tell how much more likely it is that the event A occurs than it is that it does not occur. odds of A = P(A)/P(A^c) = P(A)/(1-P(A)) If the odds are "a" it is common to say the odds are "a to 1" in favor of hypothesis

Law of Total Probability

The probability of an event is the sum of its probability across every possible condition

The multiplication rule of probability

The probability of independent events occurring in a particular sequence is equal to the product of their separate probabilities P(E1 E2 E3 ··· En) = P(E1) P(E2|E1 )P(E3|E1E2) ··· P(En|E1 ··· En−1)

3.64 Independent flips of a coin that lands on heads with probability p are made. What is the probability that the first four outcomes are (c) What is the probability that the pattern T, H, H, H occurs before the pattern H, H, H, H? Hint for part (c): How can the pattern H, H, H, H occur first?

The trick is that the only possible outcome such that a HHHH pattern appears before a T HHH is a sequence start with HHHH (think about it), and this probability is p^4 . Therefore, the probability that the pattern T, H, H, H occurs before the pattern H, H, H, H is (1 − p)^4 .

3.4 What is the probability that at least one of a pair of fair dice lands on 6, given that the sum of the dice is i, i = 2, 3, ... , 12?

To solve this task, we will be using basic probability principles. We will analyze each possible sum of the dice from 2 to 12 and consider the different outcomes that would result in those sums. We will calculate the probabilities based on the number of favorable outcomes divided by the total number of possible outcomes for each sum. To find the probability that at least one of a pair of fair dice lands on 6, given that the sum of the dice is i (where i=2,3,...,12), we need to consider the possible outcomes for each sum and calculate the probability for each case.

3.30 Suppose that an ordinary deck of 52 cards is shuffled and the cards are then turned over one at a time until the first ace appears. Given that the first ace is the 20th card to appear, what is the conditional probability that the card following it is the (a) ace of spades?

All the cards are randomly drawn from a deck of 5252 cards. Considered events: A - the first ace is the 20th card B - the 21st card is the ace of spades C - the 21st card is the two of clubs =2.34%

3.65 The color of a person's eyes is determined by a single pair of genes. If they are both blue-eyed genes, then the person will have blue eyes; if they are both brown-eyed genes, then the person will have brown eyes; and if one of them is a blue-eyed gene and the other a brown-eyed gene, then the person will have brown eyes. (Because of the latter fact, we say that the brown-eyed gene is dominant over the blue-eyed one.) A newborn child independently receives one eye gene from each of its parents, and the gene it receives from a parent is equally likely to be either of the two eye genes of that parent. Suppose that Smith and both of his parents have brown eyes, but Smith's sister has blue eyes. (c) If their first child has brown eyes, what is the probabil- ity that their next child will also have brown eyes?

=1/4

3.66 Genes relating to albinism are denoted by A and a. Only those people who receive the a gene from both par- ents will be albino. Persons having the gene pair A, a are normal in appearance and, because they can pass on the trait to their offspring, are called carriers. Suppose that a normal couple has two children, exactly one of whom is an albino. Suppose that the nonalbino child mates with a person who is known to be a carrier for albinism. (a) What is the probability that their first offspring is an albino?

=1/6

3.74 There is a 50-50 chance that the queen carries the gene for hemophilia. If she is a carrier, then each prince has a 50-50 chance of having hemophilia. If the queen has had three princes without the disease, what is the proba- bility that the queen is a carrier? If there is a fourth prince, what is the probability that he will have hemophilia?

=1/9

3.40 Urn A has 5 white and 7 black balls. Urn B has 3 white and 12 black balls. We flip a fair coin. If the outcome is heads, then a ball from urn A is selected, whereas if the outcome is tails, then a ball from urn B is selected. Sup- pose that a white ball is selected. What is the probability that the coin landed tails?

=12/37

3.12 Suppose distinct values are written on each of the 3 cards which are then randomly given the designations A, B and C. Given that card A's value is less than card B's value, find the probability it is also less than card C's value.

=2/3

3.65 The color of a person's eyes is determined by a single pair of genes. If they are both blue-eyed genes, then the person will have blue eyes; if they are both brown-eyed genes, then the person will have brown eyes; and if one of them is a blue-eyed gene and the other a brown-eyed gene, then the person will have brown eyes. (Because of the latter fact, we say that the brown-eyed gene is dominant over the blue-eyed one.) A newborn child independently receives one eye gene from each of its parents, and the gene it receives from a parent is equally likely to be either of the two eye genes of that parent. Suppose that Smith and both of his parents have brown eyes, but Smith's sister has blue eyes. (a) What is the probability that Smith possesses a blue- eyed gene?

=2/3

3.66 Genes relating to albinism are denoted by A and a. Only those people who receive the a gene from both par- ents will be albino. Persons having the gene pair A, a are normal in appearance and, because they can pass on the trait to their offspring, are called carriers. Suppose that a normal couple has two children, exactly one of whom is an albino. Suppose that the nonalbino child mates with a person who is known to be a carrier for albinism. (b) What is the conditional probability that their second offspring is an albino given that their firstborn is not?

=3/20

3.60 In a class, there are 4 first-year boys, 6 first-year girls, and 6 sophomore boys. How many sophomore girls must be present if sex and class are to be independent when a student is selected at random?

=9

3.49 An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What is the probability that all of the balls selected are white? What is the conditional probability that the die landed on 3 if all the balls selected are white?

A - all of the chosen balls are white Ei​ - result of the die roll is i ∈ {1,2,3,4,5,6} Probabilities: Since the die is fair: P(Ei) = 1/6 If the die rolls i we choose a combination of i balls, among 10 black and five white balls, therefore P(A I E1) = (5 choose 1)/(15 choose 1) = 5/15 = 1/3 P(A I E2) = (5 choose 2)/(15 choose 2) = 2/21 P(A I E3) = (5 choose 3)/(15 choose 3)= 2/91 P(A I E4) = (5 choose 4)/(15 choose 4)= 1/273 P(A I E5) = (5 choose 5)/(15 choose 5)= 1/ 3003 P(A I E6) = (5 choose 6)/(15 choose 6)= 0 P(A) = 5/66 P(E4 I A) = 22/455

How do we know if events E and F are independent?

If P(E∩F)=P(E)P(F) This condition is equivalent to P(E|F) = P(E) and to P(F|E) = P(F). Thus, the events E and F are independent if knowledge of the occurrence of one of them does not affect the probability of the other. The events E1, ... , En are said to be independent if, for any subset Ei1 , ... , Eir of them, P(Ei1 ··· Eir) = P(Ei1 )··· P(Eir) For a fixed event F, P(E|F) can be considered to be a prob- ability function on the events E of the sample space.

3.30 Suppose that an ordinary deck of 52 cards is shuffled and the cards are then turned over one at a time until the first ace appears. Given that the first ace is the 20th card to appear, what is the conditional probability that the card following it is the (b) two of clubs?

All the cards are randomly drawn from a deck of 5252 cards. Considered events: A - the first ace is the 20th card B - the 21st card is the ace of spades C - the 21st card is the two of clubs =1.89%

3.6 Consider an urn containing 12 balls, of which 8 are white. A sample of size 4 is to be drawn with replacement (& without replacement). What is the conditional probabil- ity (in each case) that the first and third balls drawn will be white given that the sample drawn contains exactly 3 white balls?

In both cases P(A I B)=1/2

Bayes' Formula

Bayes' Theorem thus gives the probability of an event based on new information that is or may be related to that event. The formula also can be used to determine how the probability of an event occurring may be affected by hypothetical new information, supposing the new information will turn out to be true. Note: the denominator of Bayes' Formula is the law of total probability

Conditional Probability Formula (probability of E given F)

Conditional Probability: Probability of E given F occurred P(E I F) = P(E∩F)/P(F) If P(F)>0 From this, we can determine P(E∩F)=P(EF)= P(E I F)P(F)

Consider that a hypothesis H that is true with probability P(H), and suppose that new evidence E is introduced. What are the conditional probabilities, given the evidence E, that H is true and that H is not true are respectively? What are the odds of H given new evidence E?

Conditional probability of H given E: P(H|E) = (P(E|H) P(H))/ P(E) Conditional probability of the compliment of H (H is not true) given E: P(H^c|E) = (P(E|H^c)P(H^c)) / P(E) New odds of H given new evidence E: P(H|E)/P(H^c|E) = P(H)P(E|H) / P(H^c)P(E|H^c)

3.28 Suppose that 5 percent of men and 0.25 percent of women are color blind. A color-blind person is chosen at random. What is the probability of this person being male? Assume that there are an equal number of males and females. What if the population consisted of twice as many males as females?

Considered events: C - a randomly chosen person is color blind M - a randomly chosen person is male F - a randomly chosen person is female I. 20/21 II. 40/41

3.34 A family has j children with probability pj, where p1 = .1, p2 = .25, p3 = .35, p4 = .3. A child from this fam- ily is randomly chosen. Given that this child is the eldest child in the family, find the conditional probability that the family has (a) only 1 child;

Ei​ - the family has i children, i ∈ {1,2,3,4} E - the chosen child is the eldest Y - the chosen child is the youngest P(E1 I E)= .24

3.34 A family has j children with probability pj, where p1 = .1, p2 = .25, p3 = .35, p4 = .3. A child from this fam- ily is randomly chosen. Given that this child is the eldest child in the family, find the conditional probability that the family has (b) 4 children.

Ei​ - the family has i children, i ∈ {1,2,3,4} E - the chosen child is the eldest Y - the chosen child is the youngest P(E4 I E)= .18

3.45 There are 3 coins in a box. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of the time. When one of the 3 coins is selected at random and flipped, it shows heads. What is the probability that it was the two-headed coin?

Events: C1​ - the coin with two heads is chosen C2​ - the fair coin is chosen C3​ - the third coin is chosen H - coin flip results in heads =4/9

3.33 Ms. Aquina has just had a biopsy on a possibly can- cerous tumor. Not wanting to spoil a weekend family event, she does not want to hear any bad news in the next few days. But if she tells the doctor to call only if the news is good, then if the doctor does not call, Ms. Aquina can conclude that the news is bad. So, being a student of prob- ability, Ms. Aquina instructs the doctor to flip a coin. If it comes up heads, the doctor is to call if the news is good and not call if the news is bad. If the coin comes up tails, the doctor is not to call. In this way, even if the doctor doesn't call, the news is not necessarily bad. Let α be the proba- bility that the tumor is cancerous; let β be the conditional probability that the tumor is cancerous given that the doc- tor does not call. (a) Which should be larger, α or β? (b) Find β in terms of α, and prove your answer in part (a).

Events: G - the news is good G=B^c - the news is bad H - the coin toss resulted in heads T=H^c - the coin toss resulted in tails C - the doctor calls a) β should be larger b) β ≥ α ⇔ P(B∣C^c) ≥ P(B) The doctor does not call if the news is bad, so every outcome when the news is bad is in the outcomes where the doctor does not call. Since this conditioning reduces the outcome space, and retains all events where the news is bad, the conditional probability is greater.

3.32 Consider two boxes, one containing 1 black and 1 white marble, the other 2 black and 1 white marble. A box is selected at random, and a marble is drawn from it at random. What is the probability that the marble is black? What is the probability that the first box was the one selected given that the marble is white?

Events: I - the first box is chosen II=I^c - the second box is chosen W - a white marble is chosen B=W^c - a black marble is chosen Since the boxes are randomly chosen: P(I)=P(II)=1/2 Since the boxes contents are known: P(W∣I)=1/2 , ​P(B∣I)=1/2, P(W∣II)=1/3, P(B∣II)=2/3 ​​P(B) can be obtained by conditioning on whether I or II happened: P(B)=P(B∣I)P(I)+P(B∣II)P(II)​ P(B)= (1/2)(1/2) + (2/3)(1/2) = 7/12 P(I IW) = 3/5

3.26 Each of 2 balls is painted either black or gold and then placed in an urn. Suppose each ball is colored black with a probability 1/2 and these events are independent a) suppose you obtain info that the gold paint has been used (and thus one of the balls is painted gold). Compute the conditional probability that both balls are painted gold. b) suppose now that the urn tips over and 1 ball falls out. It is painted gold. What is the probability that both balls are gold in this case

Given, 2 balls, painted either gold or black. Probability of each event = 1/2. Sample space =(GG,GB,BG,BB) a) P(both gold)/P(at lease 1 gold)- P(GG)/P(GG GB BG) = (1/4)/(3/4) = 1/3 b) P(both gold)/P(gold ball came out)=(1/4)/(1/2)= 1/2 This is different from a) because in a) we know that one of the three events happened, and here there are only two possible outcomes - the color of the second ball.

3.29 All the workers at a certain company drive to work and park in the company's lot. The company is interested in estimating the average number of workers in a car. Which of the following methods will enable the company to estimate this quantity? Explain your answer. 1. Randomly choose n workers, find out how many were in the cars in which they were driven, and take the aver- age of the n values. 2. Randomly choose n cars in the lot, find out how many were driven in those cars, and take the average of the n values.

In the first option we choose n workers and find out number of people in the cars traveled by them. So there may be a chance of some people travelling in the same car and there by we end up counting the same car more than once. However in the second case we are choosing n cars(In this case there is no chance of repeating count of passengers in a car) which gives us the average of n different values. So this is better way of estimating average number of passengers travelling in a car. Here in terms of probability we can say that we can probability of events is just the sum of probabilities of individual events when they are independent. Choosing n cars is independent as the count of passengers in each car are independent however choosing n persons is not independent as the count of passengers in the cars traveled by them may be dependent in the case if they travel in same car. The second option

3.48 In any given year, a male automobile policyholder will make a claim with probability pm and a female pol- icyholder will make a claim with probability pf , where pf is unequal to pm. The fraction of the policyholders that are male is α, 0 < α < 1. A policyholder is randomly chosen. If Ai denotes the event that this policyholder will make a claim in year i, show that P(A2|A1) > P(A1) Give an intuitive explanation of why the preceding inequality is true.

Intuitively, Probability of an accident for a person of an unknown gender is somewhere between the probabilites for men and women. It is an weighted average of pm​ and pf​, with weights being the probabilities that a person is male or female. If an accident occurs, the person will be more likely to be in the category with higher risk of accidents, If an accident occurs, the person will be more likely to be in the category with higher risk of accidents, so the probability that another accident happens is greater

How to compute the probability that E has occurred by conditioning on whether or not F has occurred?

Let E and F be events. We may express E as E = EF ∪ EF^c because, in order for an outcome to be in E, it must either be in both E and F or be in E but not in F. (See Figure 3.1.) As EF and EF^c are clearly mutually exclusive, we have, by Axiom 3, P(E ) = P(EF) + P(EF^c) = P(E|F)P(F) + P(E|F^c)P(F^c) = P(E|F )P(F ) + P(E|F^c)[1 − P(F)] This is an extremely useful formula, because its use often enables us to determine the probability of an event by first "conditioning" upon whether or not some second event has occurred.

3.2 If two fair dice are rolled, what is the conditional prob- ability that the first one lands on 6 given that the sum of the dice is i? Compute for all values of i between 2 and 12.

Let Ei​ be the event where the sum of the dice is i, 2≤i≤12, and F the event where the first die falls on a 6. Then the probability P(F ∣ Ei​ ), (that is the first one lands on a 6 given the sum of the dice is i, is given by) P(F ∣ Ei​)=P(Ei​ F)​/P(Ei​)

3.25 20 % of person B's phone calls are with her daughter. 65% of the time B speaks with her daughter she hangs up the phone with a smile on her face, we are interested in the conditional probability that the phone call was with her daughter. Do we have enough info to determine this probability. If yes, what is it? If no, what additional info is needed?

Not enough info. We need the probability that B hangs up the phone with a smile on her face given that the phone call was not with her daughter

3.11 Two cards are randomly chosen without replacement from an ordinary deck of 52 cards. Let B be the event that both cards are aces, let As be the event that the ace of spades is chosen, and let A be the event that at least one ace is chosen. Find (a) P(B|As) (b) P(B|A)

P(B I As) = 1/17 (B|A) = 1/33

3.14 Suppose that an ordinary deck of 52 cards (which contains 4 aces) is randomly divided into 4 hands of 13 cards each. We are interested in determining p, the prob- ability that each hand has an ace. Let Ei be the event that the ith hand has exactly one ace. Determine p = P(E1E2E3E4) by using the multiplication rule.

P(E 1​) = (4 choose 1)(48 choose 12)/(52 choose 13) P(E2 I E1) = (3 choose 1)(36 choose 12)/(39 choose 13) P(E3 I E1E2) = (2 choose 1)(24 choose 12)/(26 choose 13) P(E4 I E1E2E3) = (1 choose 1)(12 choose 12)/(13 choose 13) = 1 =.1055

3.46 Three prisoners are informed by their jailer that one of them has been chosen at random to be executed and the other two are to be freed. Prisoner A asks the jailer to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information because he already knows that at least one of the two will go free. The jailer refuses to answer the question, pointing out that if A knew which of his fellow prisoners were to be set free, then his own probability of being executed would rise from 1/3 to 1/2 because he would then be one of two prisoners. What do you think of the jailer's reasoning?

Without loss of generality, let us assume that jailer told A that B is going to be freed, then this is the probability to be analysed. If this turns out to be 1/3 then the jailer is wrong and if it is 1/2 then the jailer is correct. P(A died ∣ told B freed) Using bayes rule: P(X|Y)P(Y) = P(Y|X)P(X) = P(XY) = P(told B freed ∣ A died)P(A died)​ / P(told B freed) We can expand the denominator as a sum of disjoint probabilities = P(told B freed∣ A died)P(A died) / P(told B freed, A died)+P(told B freed, B died)+P(told B freed, C died) Now we know that unconditional probabilities of either of A, B, or C dying are same and equal to 1/3. P(W|A): If A is going to die then A will be told that B is to be freed with probability 1/2 P(W|B): If B is going to die then A will be told that B is to be freed with probability 0 P(W|C): If C is going to die then A will be told that B is to be freed with probability 1 (1/2)(1/3) / (1/2)(1/3) + 0(1/3) + 1(1/3) =1/3 The jailer is wrong, the prisoner is right

3.68 A and B are involved in a duel. The rules of the duel are that they are to pick up their guns and shoot at each other simultaneously. If one or both are hit, then the duel is over. If both shots miss, then they repeat the process. Suppose that the results of the shots are independent and that each shot of A will hit B with probability pA, and each shot of B will hit A with probability pB. What is (c) the probability that the duel ends after the nth round of shots?

[(1−pA​)(1−pB​)]^(n−1) (pA​+pB​−pA​pB​)

3.18 In a certain community, 36 percent of the families own a dog and 22 percent of the families that own a dog also own a cat. In addition, 30 percent of the families own a cat. What is (a) the probability that a randomly selected family owns both a dog and a cat? (b) the conditional probability that a randomly selected family owns a dog given that it owns a cat?

a) .0792 b) .264

3.19 A total of 46 percent of the voters in a certain city classify themselves as Independents, whereas 30 percent classify themselves as Liberals and 24 percent say that they are Conservatives. In a recent local election, 35 percent of the Independents, 62 percent of the Liberals, and 58 percent of the Conservatives voted. A voter is chosen at random. Given that this person voted in the local election, what is the probability that he or she is (a) an Independent? (b) a Liberal? (c) a Conservative? (d) What percent of voters participated in the local election?

a) .3311 b) .3826 c) .2863 d) .4862

3.53 A worker has asked her supervisor for a letter of recommendation for a new job. She estimates that there is an 80 percent chance that she will get the job if she receives a strong recommendation, a 40 percent chance if she receives a moderately good recommendation, and a 10 percent chance if she receives a weak recommendation. She further estimates that the probabilities that the rec- ommendation will be strong, moderate, and weak are .7, .2, and .1, respectively. (a) How certain is she that she will receive the new job offer? (b) Given that she does receive the offer, how likely should she feel that she received a strong recommenda- tion? a moderate recommendation? a weak recommenda- tion? (c) Given that she does not receive the job offer, how likely should she feel that she received a strong recommen- dation? a moderate recommendation? a weak recommen- dation?

a) .65 b) .86, .123, .0154 c) .4, .343, .257

3.39 (a) A gambler has a fair coin and a two-headed coin in his pocket. He selects one of the coins at random; when he flips it, it shows heads. What is the probability that it is the fair coin? (b) Suppose that he flips the same coin a second time and, again, it shows heads. Now what is the probability that it is the fair coin? (c) Suppose that he flips the same coin a third time and it shows tails. Now what is the probability that it is the fair coin?

a) 1/3 b) 1/5 c) 1