STAT 516 Answers Homework 3 January 29, 2008 Solutions by Mark Daniel Ward PROBLEMS

STAT 516 Answers Homework 3 January 29, 2008 Solutions by Mark Daniel Ward PROBLEMS 4. Write E for the event that at least one of a pair of fair dice lands on 6. Write F i for the event that the sum of the dice is i (for 2 i 12. Then P(E F i P(E F i P(F i We observe that P(E F i 0 for 2 i 6, because if at least one die lands on 6, then the sum of the dice must be at least 7. So P(E F i 0 for 2 i 6 For the values of i with i 7, we use P(E F i P(E F i P(F i to compute P(E F 7 2/36 6/36 P(E F 8 2/36 5/36 P(E F 9 2/36 4/36 P(E F 10 2/36 3/36 P(E F 11 2/36 2/36 P(E F 12 1/36 1/36 6. We are given that the 4 balls drawn contain exactly 3 white balls. Focusing attention on these 4 balls, there are ( 4 4 possible configurations (we use N to denote a non-white ball, namely WWWN, WWNW, WNWW, or NWWW. Two of these four ways have a white ball in the 1st and 3rd positions. So the desired probability is 2/4 1/2. It actually does not matter at all whether the balls were drawn with or without replacement, because we are told what the results of the selection are! 14b. We write B i for the event that the ith ball drawn is black, and W i for the event that the ith ball drawn is white. We first compute P(B 1 B 2 W 3 W 4 for an example. Note that P(B 1 (7 1 2 1 7/12. Given that B 1 occurs, then there are 5 white and 9 black balls in the urn. So P(B 2 B 1 ( 9 1 4 1 9/14. Given that B 1 and B 2 occur, then there are 5 white and 11 black balls in the urn. So P(W 3 B 1 B 2 (5 1 6 1 5/16. Given that B 1 and B 2 and W 3 occur, then there are 7 white and 11 black balls in the urn. So P(W 4 B 1 B 2 W 3 17 8 1 7/18. So P(B 1 B 2 W 3 W 4 is exactly ( ( 7 9 5 7 P(B 1 P(B 2 B 1 P(W 3 B 1 B 2 P(W 4 B 1 B 2 W 3 35/768 12 14 16 18 In a similar way, we compute ( ( 7 5 9 7 P(B 1 W 2 B 3 W 4 35 12 14 16 18 768 1

2 ( ( 7 5 7 9 P(B 1 W 2 W 3 B 4 12 14 16 18 ( ( 5 7 9 7 P(W 1 B 2 B 3 W 4 12 14 16 18 ( ( 5 7 7 9 P(W 1 B 2 W 3 B 4 12 14 16 18 ( ( 5 7 7 9 P(W 1 W 2 B 3 B 4 12 14 16 18 35 768 35 768 35 768 35 768 So the desired probability is (6(35/768 210/768 35/128. 16. Write E for the event that a baby survives, and F for the event that the baby is delivered by C section. Then P(E P(E F + P(E F c P(F c We are given P(E.98. We are also given.15, so P(F c.85. We are also given P(E F.96. Thus.98 (.96(.15 + P(E F c (.85 Solving for P(E F c gives the desired probability of P(E F c.9835. 24a. There are three equally-likely events, namely: both balls are gold, or the first ball painted was black and the second painted was gold, or the first ball painted was gold and the second painted was black. So the desired probability is 1/3. As an alternative method of solution, write E 1 for the event that the first ball painted was gold, and write E 2 for the event that the second ball painted was gold. Then we compute P(E 1 E 2 E 1 E 2 P((E 1 E 2 (E 1 E 2 P(E 1 E 2 P(E 1 E 2 P(E 1 E 2 1/4 3/4 1/3 24b. There are four equally-likely events, namely (1 both balls are gold, and the first ball to roll out was the first ball that was painted; (2 both balls are gold, and the first ball to roll out was the second ball that was painted; ( the first ball painted was black, and the second painted was gold, and the first ball to roll out was gold; (4 the second ball painted was black, and the first painted was gold, and the first ball to roll out was gold. So the desired probability is 2/4 1/2. As an alternative method of solution, write F for the event that the first ball removed is gold. The desired probability is We note that P(E 1 E 2 F P(E 1 E 2 F P(E 1 E 2 F P(F E 1 E 2 P(E 1 E 2 (1(1/4

3 and P(F E 1 E 2 P(E 1 E 2 + P(F E 1 E c 2 P(E 1 E c 2 So the desired probability is + P(F E c 1 E 2P(E c 1 E 2 + P(F E c 1 Ec 2 P(Ec 1 Ec 2 (1(1/4 + (1/2(1/4 + (1/2(1/4 + (0(1/4 1/2 P(E 1 E 2 F P(E 1 E 2 F 1/4 1/2 1/2 30. We refer to the box with 1 black and 1 white marble as Box 1. We refer to the box with 2 black and 1 white marble as Box 2. Write B i for the probability that the ith box was selected. Write E for the event that the selected marble is black; so E c is the event that the selected marble is white. First, the probability that the selected marble is black is ( 1 1 2 1 P(E P(E B 1 P(B 1 + P(E B 2 P(B 2 + 7/12 2 2 3 2 Next, the probability that the first box was the one selected, given that the marble is white, is 2 2 P(B 1 E c P(B 1 E c P(E c P(Ec B 1 P(B 1 1 P(E 1 7 12 3/5 32a. We write E 1, E 2, E 3, E 4, for the events (respectively that the family has 1, 2, 3, or 4 children. We write F for the event that the selected child is the eldest in the family. Then P(E 1 F P(E 1 F P(F E 1 P(E 1 P(F E 1 P(E 1 + P(F E 2 P(E 2 + P(F E 3 P(E 3 + P(F E 4 P(E 4 (1(.1 (1(.1 + (1/2(.25 + (1/(.35 + (1/4(..24 32b. Using the same notation, we compute P(E 4 F P(E 4 F P(F E 4 P(E 4 P(F E 1 P(E 1 + P(F E 2 P(E 2 + P(F E 3 P(E 3 + P(F E 4 P(E 4 (1/4(. (1(.1 + (1/2(.25 + (1/(.35 + (1/4(..18 Of course, the answers are still the same if we replace F by the event that the selected child is the youngest in the family.

4 37a. Write H 1 for the event that the chosen coin shows heads on the first toss. Write F for the event that the fair coin is selected. Then P(F H 1 P(F H 1 P(H 1 P(H 1 F P(H 1 F + P(H 1 F c P(F c 2 2 2 2 + (1 1 1/3 2 37b. Write H 2 for the event that the chosen coin shows heads on the first toss. Then P(F H 1 H 2 P(F H 1 H 2 P(H 1 H 2 P(H 1 H 2 F P(H 1 H 2 F + P(H 1 H 2 F c P(F c 2 4 1 4 2 + (1 1 1/5 2 37c. Given that a tail appears on the third toss, then the probability that it is the twoheaded coin is 0, so the probability that it is the fair coin is 1 in this case. (A two-headed coin will never shows tails! 40a. We use W i to denote the event that the ith ball is white and R i to denote the event that the ith ball is red. The probability that the sample contains 0 white balls is P(R 1 R 2 R 3 P(R 1 P(R 2 R 1 P(R 3 R 1 R 2 (7/12(8/1(9/14 3/13 40b. The probability that the sample contains 1 white ball is P(R 1 R 2 W 3 + P(R 1 W 2 R 3 + P(W 1 R 2 R 3 P(R 1 P(R 2 R 1 P(W 3 R 1 R 2 + P(R 1 P(W 2 R 1 P(R 3 R 1 W 2 + P(W 1 P(R 2 W 1 P(R 3 W 1 R 2 (7/12(8/1(5/14 + (7/12(5/1(8/14 + (5/12(7/1(8/14 5/13 40c. The probability that the sample contains 3 white balls is P(W 1 W 2 W 3 P(W 1 P(W 2 W 1 P(W 3 W 1 W 2 (5/12(6/1(7/14 5/52

40d. The probability that the sample contains 2 white balls can be found using the 3 answers above, writing 1 3 5 5 15/52, or a direct method of computing the probability is 13 13 52 P(W 1 W 2 R 3 + P(W 1 R 2 W 3 + P(R 1 W 2 W 3 P(W 1 P(W 2 W 1 P(R 3 W 1 W 2 + P(W 1 P(R 2 W 1 P(W 3 W 1 R 2 + P(R 1 P(W 2 R 1 P(W 3 R 1 W 2 (5/12(6/1(7/14 15/52 + (5/12(7/1(6/14 + (7/12(5/1(6/14 44. Write E 1 for the event that A is to be executed; write E 2 for the event that B is to be executed; write E 3 for the event that C is to be executed. Write F 2 for the event that the jailer says that B will be set free; write F 3 for the event that the jailer says that C will be set free. If A is to be executed, then it makes sense to assume that the jailer will choose to say that B will be set free half of the time and will say that C will be set free half of the time; i.e., if A is to be executed, the jailer will randomly choose to say that B or C will be executed. Then P(E 1 F 2 P(E 1 F 2 P(F 2 E 1 P(E 1 P(F 2 P(F 2 E 1 P(E 1 + P(F 2 E 2 P(E 2 + P(F 2 E 3 P(E 3 1 2 1 2 3 + (0 1 3 + (1 1 1/3 Similarly, we also see that P(E 1 F 3 1/3. So the prisoner is correct: The prisoner s probability of being executed is still 1/3, even if the jailer tells him the name of one of his fellow prisoners who will go free. The conditional probability does not go up to 1/2. 5 53. Write E for the event that system 1 is functioning; write F for the event that the system is functioning. Note that 1 P(F c 1 (1/2 n, since F c occurs (i.e., the system is not functioning if and only if all n of the components fail. So P(E F P(E F P(F EP(E (1(1/2 1 (1/2 n Multiplying by 2 n in the numerator and denominator yields P(E F 2n 1 2 n 1. 56. Fix n throughout the problem. Write E for the event that the nth coupon is of a new type. Write F i for the event that the nth coupon is of type i. We note that S m i1 F i, and the F i s are mutually disjoint. The probability that the nth coupon is a new type is m P(E P(E F i i1 Note that E F i occurs if each of the first n 1 types are not type of i, and the nth coupon is of type i. So P(E F i (1 p i n 1 p i. So the probability that the nth coupon is a new

6 type is P(E m (1 p i n 1 p i i1 65a. Write E for the event that the couple gives the correct answer; write F for the event that the couple agrees. There are four outcomes: 1. Write G 1 for the event that both people are correct. (Note P(G 1 (.6(.6.36. 2. Write G 2 for the event that only the wife is correct. (Note P(G 2 (.4(.6.24. 3. Write G 3 for the event that only the husband is correct. (Note P(G 3 (.6(.4.24. 4. Write G 4 for the event that neither person is correct. (Note P(G 4 (.4(.4.16. Note that the G i s are mutually disjoint. So P(E F P(E F P(G 1 P(G 1 G 4 P(G 1 P(G 1 + P(G 4.36.36 +.16 9/13 65b. Given that the couple disagrees, then they flip a coin. If the outcome of the coin corresponds to the person with the correct answer (either the wife or the husband, but not both!, then the couple has the correct answer. So the conditional probability is 1/2. (The outcome simply depends on the outcome of the coin! 67a. The probability that at least two of the components work is p 1 p 2 (1 p 3 (1 p 4 + p 1 (1 p 2 p 3 (1 p 4 + p 1 (1 p 2 (1 p 3 p 4 + (1 p 1 p 2 p 3 (1 p 4 + (1 p 1 p 2 (1 p 3 p 4 + (1 p 1 (1 p 2 p 3 p 4 + p 1 p 2 p 3 (1 p 4 + p 1 p 2 (1 p 3 p 4 + p 1 (1 p 2 p 3 p 4 + (1 p 1 p 2 p 3 p 4 + p 1 p 2 p 3 p 4 This method is much easier to write as p 1 p 2 p 3 p 4 ( i<j (1 p i (1 p j p i p j + i (1 p i + 1 p i A second method, using the complement, is to write 1 p 1 (1 p 2 (1 p 3 (1 p 4 (1 p 1 p 2 (1 p 3 (1 p 4 (1 p 1 (1 p 2 p 3 (1 p 4 (1 p 1 (1 p 2 (1 p 3 p 4 (1 p 1 (1 p 2 (1 p 3 (1 p 4 67b. The probability that at least three of the components work is p 1 p 2 p 3 (1 p 4 (1 p 5 + p 1 p 2 (1 p 3 p 4 (1 p 5 + p 1 p 2 (1 p 3 (1 p 4 p 5 + p 1 (1 p 2 p 3 p 4 (1 p 5 + p 1 (1 p 2 p 3 (1 p 4 p 5 + p 1 (1 p 2 (1 p 3 p 4 p 5 + (1 p 1 p 2 p 3 p 4 (1 p 5 + (1 p 1 p 2 p 3 (1 p 4 p 5 + (1 p 1 p 2 (1 p 3 p 4 p 5 + (1 p 1 (1 p 2 p 3 p 4 p 5 + p 1 p 2 p 3 p 4 (1 p 5 + p 1 p 2 p 3 (1 p 4 p 5 + p 1 p 2 (1 p 3 p 4 p 5 + p 1 (1 p 2 p 3 p 4 p 5 + (1 p 1 p 2 p 3 p 4 p 5 + p 1 p 2 p 3 p 4 p 5

7 This method is much easier to write as p 1 p 2 p 3 p 4 p 5 ( i<j (1 p i (1 p j p i p j + i (1 p i + 1 p i 67c. There are exactly ( n j distinct ways to pick j out of the n components; for each such selection, the probability that the j chosen components work and the other n j components each fail is exactly p j (1 p n j. So the probability that at least k of the n components are working is exactly n jk ( n j p j (1 p n j. 73. In all five parts below, there are 2 5 32 possible outcomes. 73a. There are two possibilities, namely, all are boys or all are girls. So the probability is 2/32 1/16. 73b. There is only one possibility, namely, the 3 eldest are boys and the others (i.e., the youngest two are girls. So the probability is 1/32. 73c. There are ( 5 10 distinct ways to select three of the five children as the boys; the other two children will be girls in each case. So the probability is 10/32 5/16. 73d. The two oldest children are girls, and there are 2 3 8 remaining possibilities for the others (i.e., the three youngest children. So the probability is 8/32 1/4. 73e. There is only way to have all boys, so 31 of the 32 ways have at least one girl. So the probability is 31/32. 74. Write E n for the event that A s nth roll is the last roll, i.e., A does not roll a 9 on her first n 1 tries, and B does not roll a 6 on her first n 1 tries, and A does roll a 9 on her nth try. Then P(E n 36 4 n 1 ( 1 5 n 1 ( 4 36 36 32 n 1 ( 31 n 1 ( 4 36 36 36. Notice that the E n s are disjoint, and the probability that the final roll is made by A is ( ( n 1 ( n 1 ( 32 31 4 4 1 P E n P(E n 36 36 36 36 1 ( ( 32 31 9 19 36 36 n1 n1 n1 Another method is to use conditional probabilities. On a particular move, we note that A and B both get their desired totals ( 9 for A and 6 for B with probability (4/36(5/36; in such a case, A wins. On a particular move, we note that only A (but not B gets her desired total ( 9 for A, but not 6 for B with probability (4/36(31/36; in such a case, A wins. On a particular move, we note that only B (but not A gets her desired total (not 9 for A, but 6 for B with probability (32/36(5/36; in such a case, B wins. So, given that the game is ending, the desired probability is (4/36(5/36 + (4/36(31/36 (4/36(5/36 + (4/36(31/36 + (32/36(5/36 9/19 77a. Write E 1 for the event that the first trial results in outcome 1; write F for the event that outcome 3 is the last of the three outcomes to occur. Then P(E 1 F P(E 1 F P(F E 1P(E 1 We know that P(E 1 1/3 since all three outcomes are equally likely to appear. Also, 1/3. Finally, P(F E 1 1/2; to see this, notice that we are given the appearance

8 of outcome 1 on the first roll, so the next new outcome to appear sometime in the future is equally likely to be outcome 2 or 3. Summarizing these results yields P(E 1 F (1/2(1/ 1/3 1/2 77b. Write E 2 for the event that the second trial results in outcome 1. Then P(E 1 E 2 F P((E 1 E 2 F P(F (E 1 E 2 P(E 1 E 2 We know that P(E 1 E 2 1/9 since all three outcomes are equally likely to appear each time, and the trials are independent. Also, 1/3. Finally, P(F (E 1 E 2 1/2; to see this, notice that we are given the appearance of outcome 1 on the first two rolls, so the next new outcome to appear sometime in the future is equally likely to be outcome 2 or 3. Summarizing these results yields P(E 1 E 2 F (1/2(1/9 1/3 1/6 84a. Write E for the event that A wins. Write E n for the event that A wins on her nth draw from the urn. When the balls are replaced after each draw, then the probability of E n is exactly 12 4 3(n 1 4 ( 12 2 3(n 1 1 3 8 n 1 27. The En s are disjoint, so the probability that the final roll is made by A is ( ( n 1 ( 8 1 1 1 P E n P(E n 27 3 3 1 8 9/19 27 n1 n1 n1 n1 n1 Similarly, write F for the event that B wins. Write F n for the event that B wins on her nth draw from the urn. When the balls are replaced after each draw, then the probability of F n is exactly ( 1 4 3(n 1 12 1 4 4 ( 12( 12 2 3(n 1 2 1 ( ( 8 n 1 ( 2 27 9. The Fn s are disjoint, so the probability that the final roll is made by B is ( ( n 1 ( 8 2 2 1 P F n P(F n 27 9 9 1 8 6/19 27 n1 Similarly, write G for the event that C wins. Write G n for the event that C wins on her nth draw from the urn. When the balls are replaced after each draw, then the probability of G n is exactly 12 4 3(n 1 ( 1 4 12 1 4 4 ( 12( 12 2 3(n 1 2 2 1 ( ( 3 8 n 1 ( 4 27 27. The G n s are disjoint, so the probability that the final roll is made by C is ( ( n 1 ( 8 4 4 1 P G n P(G n 27 27 27 1 8 4/19 27 n1 n1 n1 A different method of solution is available by conditional probabilities. Write a round as a sequence of three turns, by A, then B, then C. There are 3 3 27 types of rounds possible; on 27 2 3 19 of these types of rounds, at least one player draws a white ball. On 9 of these 19 types, player A gets a white ball, so the probability that A wins is 9/19. On 6 of these 19 types, player A does not get a white ball, but player B does get a white ball, so the probability that B wins is 6/19. On 4 of these 19 types, neither player A nor player B

get a white ball, but player C does get a white ball, so the probability that C wins is 4/19. 84b. If the withdrawn balls are not replaced, then the probability that A wins is P(E 4 ( 8 7 6 4 8 7 6 5 4 3 4 12 + + 7 12 11 10 9 12 11 10 9 8 7 6 15 The probability that B wins is ( 8 4 8 7 6 5 4 + 12 11 12 11 10 9 8 ( 8 7 6 5 4 3 2 4 + 12 11 10 9 8 7 6 5 53 165 The probability that C wins is ( 8 7 4 8 7 6 5 4 4 P(G + 12 11 10 12 11 10 9 8 7 ( 8 7 6 5 4 3 2 1 4 + 7 12 11 10 9 8 7 6 5 4 33 9

10 THEORETICAL EXERCISES 6. The proof is straightforward, with one application of DeMorgan s laws, and also with the observation that the independence of the E i s gives us the independence of the E c i s also. 1 P(E 1 E 2 E n P((E 1 E 2 E n c P(E1 c E2 c En c by DeMorgan s laws n P(Ei c E i s are independent, so Ei c s are also i1 n [1 P(E i ] i1 7a. We say that the urn has a homogeneous color when all of the remaining balls are the same color. At some point during the selection process, the urn becomes homogeneous. The remaining color (of the balls remaining in the homogeneous urn at this point is always exactly the color of the last ball drawn. We emphasize that each of the m + n balls are equally likely to be drawn last, and n of n m+n. these balls are white. So the desired probability is 7b. We write R i to denote the event that the red fish are the ith fish to become extinct, B i to denote the event that the blue fish are the ith fish to become extinct, and G i to denote the event that the green fish are the ith fish to become extinct. So P(R 3 is the desired probability, and we have P(R 3 P(R 1 B 2 G 3 + P(R 1 G 2 B 3 To compute P(R 1 B 2 G 3, we condition on having the green fish be extinct last, i.e., we write P(R 1 B 2 G 3 P(R 1 B 2 G 3 P(G 3 Lumping the red and blue together, we use problem 7a to see that P(G 3 g. To r+b+g compute P(R 1 B 2 G 3, we completely ignore the deaths of any green fish as the occur, and we focus attention on the red and blue fish, to find that P(R 1 B 2 G 3 b. Thus r+b P(R 1 B 2 G 3 b g. Similarly, we compute r+b r+b+g Thus, in summary, we have P(R 1 P(R 1 G 2 B 3 P(R 1 G 2 B 3 P(B 3 b g r + b r + b + g + g r + g g b r + g r + b + g b r + b + g 2rbg + bg 2 + gb 2 (r + b(r + g(r + b + g 11. Write E n for the event of tossing at least one head in the first n tosses of a coin, with probability p of landing heads each time. So P(E n 1 P(E c n 1 (1 p n We want P(E n 1/2, i.e., 1 (1 p n 1/2, or equivalently, 1 2 (1 pn. Taking the natural logarithm of both sides, ln(1/2 ln((1 p n, or equivalently, ln(1/2 n ln(1 p. Dividing by ln(1 pwhich is a negative number, we obtain ln(1/2 ln(1 p n.