M540_sol_HW3 - Chapter 2 Probability 2.1 A = cfw_FF B =...

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8 Chapter 2: Probability 2.1 A = {FF}, B = {MM}, C = {MF, FM, MM}. Then, A B = 0 / , B C = {MM}, B C = {MF, FM}, B A ={FF,MM}, C A = S , C B = C . 2.2 a. A B b. B A c. B A d. ) ( ) ( B A B A 2.3 2.4 a. b.

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Chapter 2: Probability 9 Instructor’s Solutions Manual 2.5 a. A S A B B A B A B A = = = ) ( ) ( ) ( . b. A A B B B A B B A B = = = ) ( ) ( ) ( ) ( . c. = = ) ( ) ( ) ( B B A B A B A 0 / . The result follows from part a. d. ) ( ) ( B B A B A B = = 0 / . The result follows from part b. 2.6 A = {(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (1,4), (2,4), (3,4), (4,4), (5,4), (6,4), (1,6), (2,6), (3,6), (4,6), (5,6), (6,6)} C = {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)} A B = {(2,2), (4,2), (6,2), (2,4), (4,4), (6,4), (2,6), (4,6), (6,6)} B A = {(1,2), (3,2), (5,2), (1,4), (3,4), (5,4), (1,6), (3,6), (5,6)} B A = everything but {(1,2), (1,4), (1,6), (3,2), (3,4), (3,6), (5,2), (5,4), (5,6)} A C A = 2.7 A = {two males} = {M 1 , M 2 ), (M 1 ,M 3 ), (M 2 ,M 3 )} B = {at least one female} = {(M 1 ,W 1 ), (M 2 ,W 1 ), (M 3 ,W 1 ), (M 1 ,W 2 ), (M 2 ,W 2 ), (M 3 ,W 2 ), {W 1 ,W 2 )} B = {no females} = A S B A = = B A 0 / A B A = 2.8 a. 36 + 6 = 42 b. 33 c. 18 2.9 S = {A+, B+, AB+, O+, A-, B-, AB-, O-} 2.10 a. S = {A, B, AB, O} b. P ({A}) = 0.41, P ({B}) = 0.10, P ({AB}) = 0.04, P ({O}) = 0.45. c. P ({A} or {B}) = P ({A}) + P ({B}) = 0.51, since the events are mutually exclusive. 2.11 a. Since P ( S ) = P (E 1 ) + … + P (E 5 ) = 1, 1 = .15 + .15 + .40 + 3 P (E 5 ). So, P (E 5 ) = .10 and P (E 4 ) = .20. b. Obviously, P (E 3 ) + P (E 4 ) + P (E 5 ) = .6. Thus, they are all equal to .2 2.12 a. Let L = {left tern}, R = {right turn}, C = {continues straight}. b. P (vehicle turns) = P (L) + P (R) = 1/3 + 1/3 = 2/3. 2.13 a. Denote the events as very likely (VL), somewhat likely (SL), unlikely (U), other (O). b. Not equally likely: P (VL) = .24, P (SL) = .24, P (U) = .40, P (O) = .12. c. P (at least SL) = P (SL) + P (VL) = .48. 2.14 a. P (needs glasses) = .44 + .14 = .48 b. P (needs glasses but doesn’t use them) = .14 c. P (uses glasses) = .44 + .02 = .46 2.15 a. Since the events are M.E., P ( S ) = P (E 1 ) + … + P (E 4 ) = 1. So, P (E 2 ) = 1 – .01 – .09 – .81 = .09. b. P (at least one hit) = P (E 1 ) + P (E 2 ) + P (E 3 ) = .19.
10 Chapter 2: Probability Instructor’s Solutions Manual 2.16 a. 1/3 b. 1/3 + 1/15 = 6/15 c. 1/3 + 1/16 = 19/48 d. 49/240 2.17 Let B = bushing defect, SH = shaft defect. a. P (B) = .06 + .02 = .08 b. P (B or SH) = .06 + .08 + .02 = .16 c. P (exactly one defect) = .06 + .08 = .14 d. P (neither defect) = 1 – P(B or SH) = 1 – .16 = .84 2.18 a. S = {HH, TH, HT, TT} b. if the coin is fair, all events have probability .25. c. A = {HT, TH}, B = {HT, TH, HH} d. P ( A ) = .5, P ( B ) = .75, P ( B A ) = P ( A ) = .5, P ( B A ) = P ( B ) = .75, P ( B A ) = 1. 2.19 a. (V 1 , V 1 ), (V 1 , V 2 ), (V 1 , V 3 ), (V 2 , V 1 ), (V 2 , V 2 ), (V 2 , V 3 ), (V 3 , V 1 ), (V 3 , V 2 ), (V 3 , V 3 ) b. if equally likely, all have probability of 1/9. c. A = {same vendor gets both} = {(V 1 , V 1 ), (V 2 , V 2 ), (V 3 , V 3 )} B = {at least one V2} = {(V 1 , V 2 ), (V 2 , V 1 ), (V 2 , V 2 ), (V 2 , V 3 ), (V 3 , V 2 )} So, P ( A ) = 1/3, P ( B ) = 5/9, P ( B A ) = 7/9, P ( B A ) = 1/9. 2.20 a. P ( G ) = P ( D 1 ) = P ( D 2 ) = 1/3. b. i. The probability of selecting the good prize is 1/3. ii. She will get the other dud. iii. She will get the good prize. iv. Her probability of winning is now 2/3. v. The best strategy is to switch. 2.21 P ( A ) = P ( ) ( ) ( B A B A ) = P ) ( B A + P ) ( B A since these are M.E. by Ex. 2.5. 2.22 P ( A ) = P ( ) ( B A B ) = P (B) + P ) ( B A since these are M.E. by Ex. 2.5. 2.23 All elements in B are in A , so that when B occurs, A must also occur. However, it is possible for A to occur and B not to occur. 2.24 From the relation in Ex. 2.22, P ) ( B A 0, so P ( B ) P ( A ).

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