CHAPTER 2 PROBABILITY

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1 CHAPTER 2 PROBABILITY 2. a. S = {(X,X), (X,Y), (X,Z), (Y,X), (Y,Y), (Y,Z), (Z,X), (Z,Y), (Z,Z)} b. A = {(X,X), (Y,Y), (Z,Z)} c. B = {(X,X), (X,Z), (Z,X), (Z,Z)} 4.a. A = { RRR, LLL, SSS } b. B = { RLS, RSL, LRS, LSR, SRL, SLR } c. C = { RRL, RRS, RLR, RSR, LRR, SRR } d. D = { RRL, RRS, RLR, RSR, LRR, SRR, LLR, LLS, LRL, LSL, RLL, SLL, SSR, SSL,SRS, SLS, RSS, LSS } e. D = { RRR, LLL, SSS, RLS, RSL, LRS, LSR, SRL, SLR } C D = { RRL, RRS, RLR, RSR, LRR, SRR, LLR, LLS, LRL, LSL, RLL, SLL, SSR, SSL,SRS, SLS, RSS, LSS } C D = { RRL, RRS, RLR, RSR, LRR, SRR } 6.a. b. Outcome numbers 2, 3, 5,9 c. Outcome numbers 1, 16 d. Outcome numbers 1, 2, 3, 5, 9 e. UNION:outcomes 1,2,3,5,9,16. INTERSECTION:outcome 1. f. UNION:outcomes 1, 2, 3, 5, 9, 16. INTERSECTION: b c =.

2 8.a. b. Outcomes 13, 14, 15 c. Outcomes 3, 6, 9, 12, 15 d. Outcomes 10, 11, 12, 13, 14, a. A1 A2 A3 b. A1 A2 A3 c. A1 A2 A3 d. (A1 A2 A3 ) (A1 A2 A3 ) (A1 A2 A3 )

e. A1 (A2 A3) 12.a. A = {Chev, Pont, Buick}, B = {Ford, Merc}, C = {Plym, Chrys} are three mutually exclusive events. b. No 14.a. P(A B) = 0.65 b. P((A B) ) = 0.35 c. A B`; P(A B`) =0.25 16.a. P(A B) = 0.31 b.

3 e. A1 (A2 A3) 12.a. A = {Chev, Pont, Buick}, B = {Ford, Merc}, C = {Plym, Chrys} are three mutually exclusive events. b. No 14.a. P(A B) = 0.65 b. P((A B) ) = 0.35 c. A B`; P(A B`) = a. P(A B) = 0.31 b. P(shaded region) = P(A B) - P(A B) =0.37 c. 1 - P(A B) = a. 1/6 b. P( C ranked first) =.333 c. P( C ranked first and D last) = P({CPD}) =1/6 20. P(A`) = 1 P(A) = a. {S1,C1}, {S1,C2}, {S2,C1}, {S2,C2},{S3,C1}, {S3,C2}. b. P({C1})= 0.23 c. P({S1} ) = a. P(A1 A2) = 0.3 b. P(A1 A2 ) = 0.1 c. P(exactly one) = 0.3

4 26. Since A is contained in B, then B can be written as the union of A and (B A ), two mutually exclusive events. (See diagram). From Axiom 3, P[A (B A )] = P(A) + P(B A ). Substituting P(B), P(B) = P(A) + P(B A ) or P(B) - P(A) = P(B A ). From Axiom 1, P(B A ) 0, so P(B) P(A) or P(A) P(B). For general events A and B, P(A B) P(A), and P(A B) P(A). 28. a. P(A1 ) =0.88 b. P(A1 A2 ) =0.06 c. P(A1 A2 A3 ) =0.05 d. P(at most two errors) = P(A B C) = P((A B) C) = P(A B) + P(C) P((A B) C). Using the distributive law, we may write (A B) C = (A C) (B C). Apply the addition rule to both P(A B) and this new term: P(A B C) = P(A) + P(B) P(A B) + P(C) [P(A C) + P(B C) P((A C) (B C))] = P(A) + P(B) + P(C) P(A B) P(A C) P(B C) + P(A B C). In the final step, we use the fact that (A C) (B C) = A B C. 32.a.P8,3 =336. b.c30,6 = 593,775. c.83,160 d e a = 60 b = 3 c = 36 d. # with at least one Sony = 24 e. P(at least one Sony) =0.4 P(exactly one Sony) = a. P(all from day shift) = b. P(all from same shift) = c. P(at least two shifts represented) = d. P(A1 A2 A3)= a. n1 n2 n3 = 60 runs b.10 runs.

5 40.a. P(selecting 2-75 watt bulbs) =.2967 b. P(all three are the same) =.0747 c d a.369,600 b.p(all together) = n 44.( n k )= n! = n! = ( k!(n k)! (n k)!k! n k ) The number of subsets of size k = the number of subsets of size n-k, because to each subset of size k there corresponds exactly one subset of size n-k (the n-k objects not in the subset of size k). 46. Let event A be that the individual is more than 6 feet tall. Let event B be that the individual is a professional basketball player. Then P (A B) = the probability of the individual being more than 6 feet tall, knowing that the individual is a professional basketball player, and P (B A) = the probability of the individual being a professional basketball player, knowing that the individual is more than 6 feet tall. P (A B) will be larger. Most professional BB players are tall, so the probability of an individual in that reduced sample space being more than 6 feet tall is very large. The number of individuals that are pro BB players is small in relation to the # of males more than 6 feet tall. 48.a. P(A2 A1)=0.50 b. P(A1 A2 A3 A1 )= c. P[(exactly one) (at least one)]= d. P(A 3 A1 A2)= a. P(M LS PR) = 0.05 b. P(M Pr) = 0.12 c. P(SS) = 0.44 d. P(M) = 0.49 P(Pr) =0.25 e. P(M SS Pl) =.533 f. P(SS M Pl) =.444 P(LS M Pl) = 1 - P(SS M Pl) = Let A1 be the event that #1 fails and A2 be the event that #2 fails. We assume that P(A1) = P(A2) = q and that P(A1 A2) = P(A2 A1)=r. Then one approach is as follows: P(A1 A2) = P(A2 A1) P(A1 ) = rq =.01 P(A1 A2) = P(A1 A2) + P(A1 A2) + P(A1 A2 )= rq + 2(1-r)q =.07 These two equations give 2q -.01 =.07, from which q =.04 and r =.25. Alternatively, with t =P(A1 A2) = P(A1 A2 ), t t =.07, implying t =.03 and thus q =.04 without reference to conditional probability.

6 54.a. P(A2 A1 )=0.50 b. P(A2 A3 A1 = c d P(B A) > P(B) P(B A) + P(B A) > P(B) + P(B A) P(B) + P(B A) < 1 by Exercise 55. Thus,P(B A) < 1 P(B) = P(B ) *0.3= 0.12 =P(A1 B) = P(A) P(B A).35*.6 =.21= P (A2 B).25 *.5 =.125= P( A3 B) a. P(A2 B) =.21 b. P(B) =.455 c. P(A1 B)=0.264 P(A2. B)=0.462 P(A3 B) = a. P(0 def in batch 0 def in sample) =.578 P(1 def in batch 0 def in sample) =.278 P(2 def in batch 0 def in sample) =.144 b. P(0 def in batch 1 def in sample) = 0 P(1 def in batch 1 def in sample) =.457 P(2 def in batch 1 def in sample) =.543

7 62. b. P(A B C) = c. P(B C) = d. P(C) = 0.74 e. P(A B C) = P(satis) =.51 P(mean satis) =.3922 P(median satis) =.2941 P(mode satis) =.3137 So Mean (and not Mode!) is the most likely author, while Median is least 66. Using the definition, two events A and B are independent if P(A B) = P(A); P(A B) =.6125; P(A) =.50; , so A and B are dependent. Using the multiplication rule, the events are independent if P(A B)=P(A) P(B); P(A B) =.25; P(A) P(B) = (.5)(.4) = , so A and B are dependent.

8 68. P(A1 A2) =.11, P(A1) P(A2) =.055. A1 and A2 are not independent. P(A1 A3) =.05, P(A1) P(A3) = A1 and A3 are not independent. P(A2 A3) =.07, P(A2) P(A3) =.07. A2 and A3 are independent. 70.P(two individuals match)= a. P(seam need rework) =.20 b. The desired condition is Let A1 = older pump fails, A2 = newer pump fails, and x = P(A1 A2). Then P(A1) =.10 + x, P(A2) =.05 + x, and x = P(A1 A2) = P(A1) P(A2) = (.10 + x)(.05 + x). The resulting quadratic equation, x2 -.85x = 0, has roots x =.0059 and x = Hopefully the smaller root is the actual probability of system failure. 76. p = a. P(1st detects 2nd doesn t) = 0.1 P(exactly one does)=.2 b. P(all 3 escape)= a. The events are not independent. b. P(A B) =.04. c. The numbers are quite different. 82.a. For route #1, P(late) = For route #2, P(late) =0.19 thus route #1 should be taken. b. P(4 crossing route late) = a b. 969 c d. P(best will not work) = a b a b..608 c. P(1 sent 1 received) = P(F hasn t heard after 10 times) = a. P(pass inspection) = b. P(needed no recrimping passed inspection) = P(B A) =0.926

9 96.a. P(late) =0.018 b. P(E1 on time) = a. P(all different) =.883 P(at least two the same) = b. P(at least two the same) =.476 for k=22, and =.507 for k=23 c. P(at least two have the same SS number) =0.044 Thus P(at least one coincidence ) = P(BD coincidence SS coincidence)= a. P(walks on 4th pitch) = b. P(walks on 6th) = c. P(Batter walks) = d. P(first batter scores while no one is out) = a. P(all full) =0.12 P(at least one isn t full) =0.88 b P(at least one occurs) = 1 P(none occur) = 1 (1 p1)(1 p2)(1 p3)(1 p4) P(at least two occur) = 1 P(none or 1 occur) = 1 [(1 p1)(1 p2)(1 p3)(1 p4) +p1(1 p2)(1 p3)(1 p4) +(1 p1) p2(1 p3)(1 p4) + (1 p1)(1 p2)p3(1 p4) +(1 p1)(1 p2)(1 p3)p4] 106. Let A be the event the woman carries the hemophilia gene, so that a priori P(A). Let B be the event that neither of her two sons carry the hemophilia gene. Then, using the independence suggested in the problem, P(B A) =.5 x.5 =.25. Finally, P(B A ) = 1, since a father cannot pass the hemophilia genes to his sons (it does not occur on the Y chromosomes). Therefore, P(A B) =0.2. Let C be the event that none of her three sons carry the hemophilia gene. Then P(C A) =.125 and, applying the same logic, P(A C) = = a. Obviously, a0 = 0 and a5= 1. Allan can t win without money, and he has won if he has all 5 dollars. b. By the law of total probability, a2 = P(H)P(Allan wins /w/ $3) + P(T)P(Allan wins /w/ $1) =.5a3 +.5a1 c. The same logic yields four equations: a1 =.5a2 +.5a0, a2 =.5a3 +.5a1, a3 =.5a4 +.5a2, and a4 =.5a5 +.5a3. Rewrite the last equation as 2a4 = a5 + a3, then a4 a3 = a5 a4. Rewrite the others similarly, so we see the increment between successive a s is the same. Clearly then, with the initial conditions from (a), a1 = 1/5, a2 = 2/5, a3 = 3/5, and a4 d. Following the pattern suggested in (c) the chance Allan beats Beth with their initial fortunes being $a and $b is a/(a+b). 110.P(II=D I=Aa) =0.43

CHAPTER a. S = { 1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 4123, 4132, 3214, 3241, 4213, 4231 }

CHAPTER a. S = { 1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 4123, 4132, 3214, 3241, 4213, 4231 } CHTER Section.. a. S { 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4 } b. Event contains the outcomes where is first in the list: { 4, 4, 4, 4 } c. Event B contains the outcomes where is first or second: