A First Course in Probability, 10th Edition Solution Manual

Chapter 2 Problems 1. (a) S = {(r, r), (r, g), (r, b), (g, r), (g, g), (g, b), (b, r), b, g), (b, b)} (b) S = {(r, g), (r, b), (g, r), (g, b), (b, r), (b, g)} 2. S = {(n, x1, …, xn−1), n ≥ 1, xi ≠ 6, i = 1, …, n − 1}, with the interpretation that the outcome is (n, x1, …, xn−1) if the first 6 appears on roll n, and xi appears on roll, i, i = 1, …, n − 1. The c event (∪∞ n =1 En ) is the event that 6 never appears. 3. EF = {(1, 2), (1, 4), (1, 6), (2, 1), (4, 1), (6, 1)}. E ∪ F occurs if the sum is odd or if at least one of the dice lands on 1. FG = {(1, 4), (4, 1)}. EFc is the event that neither of the dice lands on 1 and the sum is odd. EFG = FG. 4. A = {1,0001,0000001, …} B = {01, 00001, 00000001, …} (A ∪ B)c = {00000 …, 001, 000001, …} 5. (a) 25 = 32 (b) W = {(1, 1, 1, 1, 1), (1, 1, 1, 1, 0), (1, 1, 1, 0, 1), (1, 1, 0, 1, 1), (1, 1, 1, 0, 0), (1, 1, 0, 1, 0) (1, 1, 0, 0, 1), (1, 1, 0, 0, 0), (1, 0, 1, 1, 1), (0, 1, 1, 1, 1), (1, 0, 1, 1, 0), (0, 1, 1, 1, 0) (0, 0, 1, 1, 1) (0, 0, 1, 1, 0), (1, 0, 1, 0, 1)} (c) 8 (d) AW = {(1, 1, 1, 0, 0), (1, 1, 0, 0, 0)} 6. (a) S = {(1, g), (0, g), (1, f), (0, f), (1, s), (0, s)} (b) A = {(1, s), (0, s)} (c) B = {(0, g), (0, f), (0, s)} (d) {(1, s), (0, s), (1, g), (1, f)} 7. (a) 615 (b) 615 − 315 (c) 415 8. (a) .8 (b) .3 (c) 0 10 Copyright © 2018 Pearson Education, Inc. Chapter 2 9. 11 Choose a customer at random. Let A denote the event that this customer carries an American Express card and V the event that he or she carries a VISA card. P(A ∪ V) = P(A) + P(V) − P(AV) = .24 + .61 − .11 = .74. Therefore, 74 percent of the establishment’s customers carry at least one of the two types of credit cards that it accepts. 10. Let R and N denote the events, respectively, that the student wears a ring and wears a necklace. (a) P(R ∪ N) = 1 − .6 = .4 (b) .4 = P(R ∪ N) = P(R) + P(N) − P(RN) = .2 + .3 − P(RN) Thus, P(RN) = .1 11. Let A be the event that a randomly chosen person is a cigarette smoker and let B be the event that she or he is a cigar smoker. (a) 1 − P(A ∪ B) = 1 − (.07 + .28 − .05) = .7. Hence, 70 percent smoke neither. (b) P(AcB) = P(B) − P(AB) = .07 − .05 = .02. Hence, 2 percent smoke cigars but not cigarettes. 12. (a) P(S ∪ F ∪ G) = (28 + 26 + 16 − 12 − 4 − 6 + 2)/100 = 1/2 The desired probability is 1 − 1/2 = 1/2. (b) Use the Venn diagram below to obtain the answer 32/100. S 14 F 10 10 2 2 4 8 G (c) Since 50 students are not taking any of the courses, the probability that neither one is  50  100  taking a course is     = 49/198 and so the probability that at least one is taking 2  2  a course is 149/198. Copyright © 2018 Pearson Education, Inc. 12 Chapter 2 13. I 1000 (a) (b) (c) (d) (e) II 7000 19000 20,000 12,000 11,000 68,000 10,000 1000 1000 3000 0 III 14. 15. P(M) + P(W) + P(G) − P(MW) − P(MG) − P(WG) + P(MWG) = .312 + .470 + .525 − .086 − .042 − .147 + .025 = 1.057  13   52  (a) 4     5 5  4 12  4  4  4   52  (b) 13          2  3  1  1  1   5   13   4  4  44   52  (c)          2   2  2  1   5   4 12  4  4   52  (d) 13         3  2  1  1   5   4  48   52  (e) 13       4  1   5  16. (a) 6 ⋅5⋅ 4 ⋅3⋅ 2 65 5 6 ⋅ 5 ⋅ 4   3 (d) 21 (g) (b) (e) 5 6 5 ⋅ 4 ⋅ 3  2 65 5 6 ⋅ 5  3 5 6 6 65 Copyright © 2018 Pearson Education, Inc. (c) (f)  6   5  3    4     2   2  2  65 5 6 ⋅ 5   4 5 6 Chapter 2 13 17. 15   10   7       8   8   1  = .1102  25   9      16   1  18. 2 ⋅ 4 ⋅ 16 52 ⋅ 51 19. 4/36 + 4/36 +1/36 + 1/36 = 5/18 20. Let A be the event that you are dealt blackjack and let B be the event that the dealer is dealt blackjack. Then, P(A ∪ B) = P(A) + P(B) − P(AB) = 4 ⋅ 4 ⋅ 16 4 ⋅ 4 ⋅ 16 ⋅ 3 ⋅ 15 + 52 ⋅ 51 52 ⋅ 51 ⋅ 50 ⋅ 49 = .0983 where the preceding used that P(A) = P(B) = 2 × is dealt blackjack is .9017. 21. 4 ⋅ 16 . Hence, the probability that neither 52 ⋅ 51 (a) p1 = 4/20, p2 = 8/20, p3 = 5/20, p4 = 2/20, p5 = 1/20 (b) There are a total of 4 ⋅ 1 + 8 ⋅ 2 + 5 ⋅ 3 + 2 ⋅ 4 + 1 ⋅ 5 = 48 children. Hence, q1 = 4/48, q2 = 16/48, q3 = 15/48, q4 = 8/48, q5 = 5/48 22. The ordering will be unchanged if for some k, 0 ≤ k ≤ n, the first k coin tosses land heads and the last n − k land tails. Hence, the desired probability is (n + 1/2n 23. The answer is 5/12, which can be seen as follows: 1 = P{first higher} + P{second higher} + p{same} = 2P{second higher} + p{same} = 2P{second higher} + 1/6 Another way of solving is to list all the outcomes for which the second is higher. There is 1 outcome when the second die lands on two, 2 when it lands on three, 3 when it lands on four, 4 when it lands on five, and 5 when it lands on six. Hence, the probability is (1 + 2 + 3 + 4 + 5)/36 = 5/12. 25.  26  P(En) =    36  n −1 6 , 36 ∞ 2  P( En ) = 5 n =1 Copyright © 2018 Pearson Education, Inc. 14 27. Chapter 2 Imagine that all 10 balls are withdrawn P(A) = 28. 3 ⋅ 9!+ 7 ⋅ 6 ⋅ 3 ⋅ 7!+ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 5!+ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 3 ⋅ 3! 10!  5  6 8  + +  3 3 3 P{same} =        19    3  5  6  8  19  P{different} =        1  1  1   3  If sampling is with replacement P{same} = 53 + 63 + 83 (19)3 P{different} = P(RBG) + P{BRG) + P(RGB) + … + P(GBR) = 29. (a) 6⋅5⋅6⋅8 (19)3 n(n − 1) + m(m − 1) ( n + m)(n + m − 1) (b) Putting all terms over the common denominator (n + m)2(n + m − 1) shows that we must prove that n2(n + m − 1) + m2(n + m − 1) ≥ n(n − 1)(n + m) + m(m − 1)(n + m) which is immediate upon multiplying through and simplifying. 30.  7  8     3! 3 3 (a)    = 1/18  8  9     4!  4  4   7  8     3! 3 3 (b)    − 1/18 = 1/6  8  9     4!  4  4   7  8   7  8     +    3 4 4 3 (c)       = 1/2  8  9      4  4  Copyright © 2018 Pearson Education, Inc. Chapter 2 31. 15 P({complete} = P{same} = 32. g (b + g − 1)! g = + + (b g )! b g 33.  5 15      2  2  = 70 323  20    4 34.  32   52       13   13  35. 12 16 18      3 2 2 (a)      46    7  34  12  34       7 1 6 (b) 1 −   −     46   46      7 7 12  16  18   + +  7 7 7 (c)        46    7  12  34  16  30  12 16 18            3 4 3 4 3 3 1 (d) P ( R3 ∪ B3 ) = P( R3 ) + P ( B3 ) − P ( R3 B3 ) =    +    −      46   46   46        7 7 7 36.  4   52  (a)     ≈ .0045,  2  2   4   52  (b) 13     = 1/17 ≈ .0588  2  2  Copyright © 2018 Pearson Education, Inc. 16 37. Chapter 2  7  10  (a)     = 1/12 ≈ .0833 5  5   7  3  10  (b)      + 1/12 = 1/2  4  1   5  38. 3  n 1/2 =     or n(n − 1) = 12 or n = 4.  2  2 39. 5 ⋅ 4 ⋅ 3 12 = 5 ⋅ 5 ⋅ 5 25 40. .8134; .1148 41. 1− 42.  35  1−    36  43. 44. 54 64 n 2(n − 1)(n − 2) 2 = in a line n! n 2n(n − 2)! 2 = if in a circle, n ≥ 2 n! n −1 (a) If A is first, then A can be in any one of 3 places and B’s place is determined, and the others can be arranged in any of 3! ways. As a similar result is true, when B is first, we see that the probability in this case is 2 ⋅ 3 ⋅ 3!/5! = 3/10 (b) 2 ⋅ 2 ⋅ 3!/5! = 1/5 45. 46. (c) 2 ⋅ 3!/5! = 1/10 (n − 1) k −1 if do not discard 1/n if discard, nk If n in the room, P{all different} = 12 ⋅ 11 ⋅ 12 ⋅ 12 ⋅ ⋅ (13 − n) ⋅ 12 When n = 5 this falls below 1/2. (Its value when n = 5 is .3819) Copyright © 2018 Pearson Education, Inc. Chapter 2 17 47. 8 5     2   2  = .1399 14    5 48.  12  8  (20)!    4 4  4  4  (3!) (2!) 49.  6  6  12        3  3   6  50.  13  39  8  31  52  39           5  8  8  5   13  13  51. n n−m / Nn   (n − 1) m   52. (a) (12) 20 20 ⋅ 18 ⋅ 16 ⋅ 14 ⋅ 12 ⋅ 10 ⋅ 8 ⋅ 6 20 ⋅ 19 ⋅ 18 ⋅ 17 ⋅ 16 ⋅ 15 ⋅ 14 ⋅ 13  10  9  8! 6    2  1  6  2! (b) 20 ⋅ 19 ⋅ 18 ⋅ 17 ⋅ 16 ⋅ 15 ⋅ 14 ⋅ 13 53. Let Ai be the event that couple i sit next to each other. Then P (∪i4=1 Ai ) = 4 2 ⋅ 7! 22 ⋅ 6! 23 ⋅ 5! 24 ⋅ 4! −6 +4 − 8! 8! 8! 8! and the desired probability is 1 minus the preceding. 54. P(S ∪ H ∪ D ∪ C) = P(S) + P(H) + P(D) + P(C) − P(SH) − … − P(SHDC)  39   26  13  4  6  4  13 13 13 =  −  +    52   52   52         13   13   13   39   26  4  − 6  + 4 13 13 =      52     13  Copyright © 2018 Pearson Education, Inc. Chapter 2 18 55. (a) P(S ∪ H ∪ D ∪ C) = P(S) + … − P(SHDC) 3 4  2  2  2  48   2   46   2   44  4   6     4         2 2 2 9 2 7 2 5 =   −     +     −      52   52   52   52           13   13   13   13   50   48   46   44  4  − 6  + 4  −   11 9 7 5 =      52     13   48   13  44  13  40  13         9 2 5 3 1 (b) P(1 ∪ 2 ∪ … ∪ 13) =   −    +     52   52   52         13   13   13  56. Player B. If Player A chooses spinner (a) then B can choose spinner (c). If A chooses (b) then B chooses (a). If A chooses (c) then B chooses (b). In each case B wins probability 5/9. Copyright © 2018 Pearson Education, Inc. Chapter 2 19 Theoretical Exercises i =1 5. Fi = Ei ∩ E cj 6. (a) EFcGc j =1 (b) EFcG (c) E ∪ F ∪ G (d) EF ∪ EG ∪ FG (e) EFG (f) EcFcGc (g) EcFcGc ∪ EFcGc ∪ EcFGc ∪ EcFcG (h) (EFG)c (i) EFGc ∪ EFcG ∪ EcFG (j) S 8. The number of partitions that has n + 1 and a fixed set of i of the elements 1, 2, …, n as a n subset is Tn−i. Hence, (where T0 = 1). Hence, as there are   such subsets. i n n n −1 n n n       Tn+1 =    Tn −i = 1 +    Tn −i = 1 +    Tk . i i i =0   i =0   k =1  k  11. 1 ≥ P(E ∪ F) = P(E) + P(F) − P(EF) 12. P(EFc ∪ EcF) = P(EFc) + P(EcF) = P(E) − P(EF) + P(F) − P(EF) 13. E = EF ∪ EFc 15.  M  N      k  r − k  M + N     r  Copyright © 2018 Pearson Education, Inc. 20 16. Chapter 2 P(E1 … En) ≥ P(E1 … En−1) + P(En) − 1 by Bonferonni’s Ineq. n −1 ≥  P( Ei ) − (n − 2) + P(En) − 1 by induction hypothesis 1 19. 21.  n  m     (n − r + 1)  r − 1 k − r  n + m   (n + m − k + 1)  k −1  Let y1, y2, …, yk denote the successive runs of losses and x1, …, xk the successive runs of wins. There will be 2k runs if the outcome is either of the form y1, x1, …, yk xk or x1y1, … xk, yk where all xi, yi are positive, with x1 + … + xk = n, y1 + … + yk = m. By Proposition 6.1 there are  n − 1  m − 1 2   number of outcomes and so  k − 1 k − 1   n − 1 m − 1  m + n  P{2k runs} = 2     .  k − 1 k − 1   n  There will be 2k + 1 runs if the outcome is either of the form x1, y1, …, xk, yk, xk+1 or y1, x1, …, yk, xk yk + 1 where all are positive and  xi = n,  yi = m. By Proposition 6.1 there are  n − 1 m − 1  n − 1 m − 1    outcomes of the first type and    of the second.  k  k − 1   k − 1 k  Copyright © 2018 Pearson Education, Inc.