0. So they are not independent 2.82 P(B|A) = P(BโฉA)/P(A) = P(A)/P(A) = 1 P(A|B) = P(AโฉB)/P(B) = P(A)/P(B). b. 0.37 f. 1 โ 0.67 = 0.33 i. 1/.4 = 0.25 c. 0.10 d. 0.40 + 0.37 โ 0.10 = 0.67 g. 1 โ 0.10 = 0.90 Chapter 2: Probability 18 Instructorโs Solutions Manual P( A) , since A and B are M.E. events. P( A) + P( B ) 2.83 P(A | A โช B ) = P(A)/P( A โช B ) = 2.84 Note that if P( A2 โฉ A3 ) = 0, then P( A1 โฉ A2 โฉ A3 ) also equals 0. The result follows from Theorem 2.6. 2.85 P( A | B ) = P( A โฉ B )/P( B ) = P( B | A) P( A) [1 โ P( B | A)]P( A) [1 โ P( B )]P( A) = = = P( B ) P( B ) P( B ) P( B ) P( A) = P( A). So, A and B are independent. P( B ) P( A | B ) P( B ) [1 โ P( A | B )]P( B ) = . From the above, P( B | A ) = P( B โฉ A ) /P( A ) = P( A ) P( A ) [1 โ P( A)]P( B ) = P( A ) P( B ) = P( B ). So, A and B are independent. So P( B | A ) = P( A ) P( A ) A and B are independent 2.86 a. No. It follows from P( A โช B ) = P(A) + P(B) โ P(AโฉB) โค 1. b. P(AโฉB) โฅ 0.5 c. No. d. P(AโฉB) โค 0.70. 2.87 a. P(A) + P(B) โ 1. b. the smaller of P(A) and P(B). 2.88 a. Yes. b. 0, since they could be disjoint. c. No, since P(AโฉB) cannot be larger than either of P(A) or P(B). d. 0.3 = P(A). 2.89 a. 0, since they could be disjoint. b. the smaller of P(A) and P(B). 2.90 a. (1/50)*(1/50) = 0.0004. b. P(at least one injury) = 1 โ P(no injuries in 50 jumps) = 1 = (49/50)50 = 0.636. Your friend is not correct. 2.91 If A and B are M.E., P( A โช B ) = P(A) + P(B). This value is greater than 1 if P(A) = 0.4 and P(B) = 0.7. So they cannot be M.E. It is possible if P(A) = 0.4 and P(B) = 0.3. 2.92 a. The three tests are independent. So, the probability in question is (.05)3 = 0.000125. b. P(at least one mistake) = 1 โ P(no mistakes) = 1 โ (.95)3 = 0.143. 2.93 Part a is found using the Addition Rule. Parts b and c use DeMorganโs Laws. a. 0.2 + 0.3 โ 0.4 = 0.1 Chapter 2: Probability 19 Instructorโs Solutions Manual b. 1 โ 0.1 = 0.9 c. 1 โ 0.4 = 0.6 P ( A โฉ B ) P ( B ) โ P ( A โฉ B ) .3 โ .1 d. P( A | B ) = = = = 2/3. P( B ) P( B ) .3 2.94 Define the events A: device A detects smoke B: device B detects smoke a. P( A โช B ) = .95 + .90 – .88 = 0.97. b. P(smoke is undetected) = 1 – P( A โช B ) = 1 โ 0.97 = 0.03. 2.95 Let H denote a hit and let M denote a miss. Then, she wins the game in three trials with the events HHH, HHM, and MHH. If she begins with her right hand, the probability she wins the game, assuming independence, is (.7)(.4)(.7) + (.7)(.4)(.3) + (.3)(.4)(.7) = 0.364. 2.96 Using the results of Ex. 2.95: a. 0.5 + 0.2 โ (0.5)(0.2) = 0.6. b. 1 โ 0.6 = 0.4. c. 1 โ 0.1 = 0.9. 2.97 a. P(current flows) = 1 โ P(all three relays are open) = 1 โ (.1)3 = 0.999. b. Let A be the event that current flows and B be the event that relay 1 closed properly. Then, P(B|A) = P(BโฉA)/P(A) = P(B)/P(A) = .9/.999 = 0.9009. Note that B โ A . 2.98 Series system: P(both relays are closed) = (.9)(.9) = 0.81 Parallel system: P(at least one relay is closed) = .9 + .9 โ .81 = 0.99. 2.99 Given that P( A โช B ) = a, P(B) = b, and that A and B are independent. Thus P( A โช B ) = 1 โ a and P(BโฉA) = bP(A). Thus, P( A โช B ) = P(A) + b – bP(A) = 1 โ a. Solve for P(A). 2.100 P( A โช B | C ) = P(( A โช B ) โฉ C ) P(( A โฉ C ) โช ( B โฉ C )) = = P (C ) P (C ) P( A โฉ C ) + P( B โฉ C ) โ P( A โฉ B โฉ C ) = P(A|C) + P(B|C) + P(AโฉB|C). P (C ) 2.101 Let A be the event the item gets past the first inspector and B the event it gets past the second inspector. Then, P(A) = 0.1 and P(B|A) = 0.5. Then P(AโฉB) = .1(.5) = 0.05. 2.102 Define the events: I: disease I us contracted P(I) = 0.1, P(II) = 0.15, and P(IโฉII) = 0.03. a. P(I โช II) = .1 + .15 โ .03 = 0.22 b. P(IโฉII|I โช II) = .03/.22 = 3/22. II: disease II is contracted. Then, 2.103 Assume that the two state lotteries are independent. a. P(666 in CT|666 in PA) = P(666 in CT) = 0.001 b. P(666 in CTโฉ666 in PA) = P(666 in CT)P(666 in PA) = .001(1/8) = 0.000125. Chapter 2: Probability 20 Instructorโs Solutions Manual 2.104 By DeMorganโs Law, P( A โฉ B ) = 1 โ P( A โฉ B ) = 1 โ P( A โช B ) . Since P( A โช B ) โค P( A ) + P( B ) , P( A โฉ B ) โฅ 1 โ P( A ) โ P( B ). 2.105 P(landing safely on both jumps) โฅ โ 0.05 โ 0.05 = 0.90. 2.106 Note that it must be also true that P( A ) = P( B ) . Using the result in Ex. 2.104, P( A โฉ B ) โฅ 1 โ 2 P( A ) โฅ 0.98, so P(A) โฅ 0.99. 2.107 (Answers vary) Consider flipping a coin twice. Define the events: A: observe at least one tail B: observe two heads or two tails C: observe two heads 2.108 Let U and V be two events. Then, by Ex. 2.104, P(U โฉ V ) โฅ 1 โ P(U ) โ P(V ). Let U = AโฉB and V = C. Then, P( A โฉ B โฉ C ) โฅ 1 โ P( A โฉ B ) โ P(C ) . Apply Ex. 2.104 to P( A โฉ B ) to obtain the result. 2.109 This is similar to Ex. 2.106. Apply Ex. 2.108: 0.95 โค 1 โ P( A ) โ P( B ) โ P(C ) โค P( A โฉ B โฉ C ) . Since the events have the same probability, 0.95 โค 1 โ 3P( A ) . Thus, P(A) โฅ 0.9833. 2.110 Define the events: I: item is from line I II: item is from line II N: item is not defective Then, P(N) = P( N โฉ ( I โช II )) = P(NโฉI) + P(NโฉII) = .92(.4) + .90(.6) = 0.908. 2.111 Define the following events: A: buyer sees the magazine ad B: buyer sees the TV ad C: buyer purchases the product The following are known: P(A) = .02, P(B) = .20, P(AโฉB) = .01. Thus P( A โฉ B ) = .21. Also, P(buyer sees no ad) = P( A โฉ B ) = 1 โ P( A โช B ) = 1 โ 0.21 = 0.79. Finally, it is known that P(C | A โช B ) = 0.1 and P(C | A โฉ B ) = 1/3. So, we can find P(C) as P(C) = P(C โฉ ( A โช B )) + P(C โฉ ( A โฉ B )) = (1/3)(.21) + (.1)(.79) = 0.149. 2.112 a. P(aircraft undetected) = P(all three fail to detect) = (.02)(.02)(.02) = (.02)3. b. P(all three detect aircraft) = (.98)3. 2.113 By independence, (.98)(.98)(.98)(.02) = (.98)3(.02). 2.114 Let T = {detects truth} and L = {detects lie}. The sample space is TT, TL, LT, LL. Since one suspect is guilty, assume the guilty suspect is questioned first: a. P(LL) = .95(.10) = 0.095 b. P(LT) = ..95(.9) = 0.885 Chapter 2: Probability 21 Instructorโs Solutions Manual b. P(TL) = .05(.10) = 0.005 d. 1 โ (.05)(.90) = 0.955 2.115 a. From the description of the problem, there is a 50% chance a car will be rejected. To find the probability that three out of four will be rejected (i.e. the drivers chose team 2), โ4โ note that there are โโ โโ = 4 ways that three of the four cars are evaluated by team 2. Each โ 3โ one has probability (.5)(.5)(.5)(.5) of occurring, so the probability is 4(.5)4 = 0.25. b. The probability that all four pass (i.e. all four are evaluated by team 1) is (.5)4 = 1/16. 2.116 By the complement rule, P(system works) = 1 โ P(system fails) = 1 โ (.01)3. 2.117 By independence, (.75)(.75)(.75)(.75) = (.75)4. 2.118 If the victim is to be saved, a proper donor must be found within eight minutes. The patient will be saved if the proper donor is found on the 1st, 2nd, 3rd, or 4th try. But, if the donor is found on the 2nd try, that implies he/she wasnโt found on the 1st try. So, the probability of saving the patient is found by, letting A = {correct donor is found}: P(save) = P(A) + P( A A) + P( A A A) + P( A A A A) . By independence, this is .4 + .6(.4) + (.6)2(.4) + (.6)3(.4) = 0.8704 2.119 a. Define the events: A: obtain a sum of 3 B: do not obtain a sum of 3 or 7 Since there are 36 possible rolls, P(A) = 2/36 and P(B) = 28/36. Obtaining a sum of 3 before a sum of 7 can happen on the 1st roll, the 2nd roll, the 3rd roll, etc. Using the events above, we can write these as A, BA, BBA, BBBA, etc. The probability of obtaining a sum of 3 before a sum of 7 is given by P(A) + P(B)P(A) + [P(B)]2P(A) + [P(B)]3P(A) + โฆ . (Here, we are using the fact that the rolls are independent.) This is an infinite sum, and it follows as a geometric series. Thus, 2/36 + (28/36)(2/36) + (28/36)2(2/26) + โฆ = 1/4. b. Similar to part a. Define C: obtain a sum of 4 D: do not obtain a sum of 4 or 7 Then, P(C) = 3/36 and P(D) = 27/36. The probability of obtaining a 4 before a 7 is 1/3. 2.120 Denote the events G: good refrigerator D: defective refrigerator a. If the last defective refrigerator is found on the 4th test, this means the first defective refrigerator was found on the 1st, 2nd, or 3rd test. So, the possibilities are DGGD, GDGD, and GGDD. So, the probability is ( 62 )( 45 )( 43 ) 13 . The probabilities associated with the other two events are identical to the first. So, the desired probability is 3 ( 62 )( 45 )( 43 ) 13 = 15 . b. Here, the second defective refrigerator must be found on the 2nd, 3rd, or 4th test. Define: A1: second defective found on 2nd test A2: second defective found on 3rd test A3: second defective found on 4th test Clearly, P(A1) = ( 62 )( 15 ) = 151 . Also, P(A3) = 15 from part a. Note that A2 = {DGD, GDD}. Thus, P(A2) = 2 ( 62 )( 45 )( 14 ) = 152 . So, P(A1) + P(A2) + P(A3) = 2/5. c. Define: B1: second defective found on 3rd test Chapter 2: Probability 22 Instructorโs Solutions Manual B2: second defective found on 4th test Clearly, P(B1) = 1/4 and P(B2) = (3/4)(1/3) = 1/4. So, P(B1) + P(B2) = 1/2. 2.121 a. 1/n b. nnโ1 โ n1โ1 = 1/n. nnโ1 โ nnโโ12 โ nโ1 2 = 1/n. c. P(gain access) = P(first try) + P(second try) + P(third try) = 3/7. 2.122 Applet exercise (answers vary). 2.123 Applet exercise (answers vary). 2.124 Define the events for the voter: D: democrat R: republican P( F | D ) P( D ) .7(.6) P( D | F ) = = = 7/9 P( F | D ) P( D ) + P( F | R ) P( R ) .7(.6) + .3(.4) F: favors issue 2.125 Define the events for the person: D: has the disease H: test indicates the disease Thus, P(H|D) = .9, P( H | D ) = .9, P(D) = .01, and P( D ) = .99. Thus, P( H | D ) P( D ) P( D | H ) = = 1/12. P( H | D ) P( D ) + P( H | D ) P( D ) 2.126 a. (.95*.01)/(.95*.01 + .1*.99) = 0.08756. b. .99*.01/(.99*.01 + .1*.99) = 1/11. c. Only a small percentage of the population has the disease. d. If the specificity is .99, the positive predictive value is .5. e. The sensitivity and specificity should be as close to 1 as possible. 2.127 a. .9*.4/(.9*.4 + .1*.6) = 0.857. b. A larger proportion of the population has the disease, so the numerator and denominator values are closer. c. No; if the sensitivity is 1, the largest value for the positive predictive value is .8696. d. Yes, by increasing the specificity. e. The specificity is more important with tests used for rare diseases. 2.128 For i = 1, 2, 3, let Fi represent the event that the plane is found in region i and Ni be the complement. Also Ri is the event the plane is in region i. Then P(Fi|Ri) = 1 โ ฮฑi and P(Ri) = 1/3 for all i. Then, ฮฑ 1 13 P( N 1 | R1 ) P( R1 ) a. P( R1 | N 1 ) = = P( N 1 | R1 ) P( R1 ) + P( N 1 | R2 ) P( R2 ) + P( N 1 | R3 ) P( R3 ) ฮฑ 1 13 + 13 + 13 = ฮฑ1 ฮฑ1 + 2 . b. Similarly, P( R2 | N 1 ) = 1 and ฮฑ1 + 2 c. P( R3 | N 1 ) = 1 . ฮฑ1 + 2 Chapter 2: Probability 23 Instructorโs Solutions Manual 2.129 Define the events: P: positive response M: male respondent F: female respondent P(P|F) = .7, P(P|M) = .4, P(M) = .25. Using Bayesโ rule, P( P | M ) P( M ) .6(.25) P( M | P ) = = 0.4. = P( P | M ) P( M ) + P( P | F ) P( F ) .6(.25) + .3(.75) 2.130 Define the events: C: contract lung cancer S: worked in a shipyard Thus, P(S|C) = .22, P( S | C ) = .14, and P(C) = .0004. Using Bayesโ rule, P( S | C ) P(C ) .22(.0004) P(C | S ) = = 0.0006. = P( S | C ) P(C ) + P( S | C ) P(C ) .22(.0004) + .14(.9996) 2.131 The additive law of probability gives that P( AฮB ) = P( A โฉ B ) + P( A โฉ B ) . Also, A and B can be written as the union of two disjoint sets: A = ( A โฉ B ) โช ( A โฉ B ) and B = ( A โฉ B ) โช ( A โฉ B ) . Thus, P( A โฉ B ) = P( A) โ P( A โฉ B ) and P( A โฉ B ) = P( B ) โ P( A โฉ B ) . Thus, P( AฮB ) = P( A) + P( B ) โ 2 P( A โฉ B ) . 2.132 a. Let P( A | B ) = P( A | B ) = p. By the Law of Total Probability, P( A) = P( A | B ) P( B ) + P( A | B ) P( B ) = p (P( B ) + P( B ) ) = p. Thus, A and B are independent. b. P( A) = P( A | C ) P(C ) + P( A | C ) P(C ) > P( B | C ) P(C ) + P( B | C ) P(C ) = P( B ) . 2.133 Define the events: G: student guesses C: student is correct P(C | G ) P(G ) 1(.8) = P(G | C ) = = 0.9412. P(C | G ) P(G ) + P(C | G ) P(G ) 1(.8) + .25(.2) 2.134 Define F as โfailure to learn. Then, P(F|A) = .2, P(F|B) = .1, P(A) = .7, P(B) = .3. By Bayesโ rule, P(A|F) = 14/17. 2.135 Let M = major airline, P = private airline, C = commercial airline, B = travel for business a. P(B) = P(B|M)P(M) + P(B|P)P(P) + P(B|C)P(C) = .6(.5) + .3(.6) + .1(.9) = 0.57. b. P(BโฉP) = P(B|P)P(P) = .3(.6) = 0.18. c. P(P|B) = P(BโฉP)/P(B) = .18/.57 = 0.3158. d. P(B|C) = 0.90. 2.136 Let A = womanโs name is selected from list 1, B = womanโs name is selected from list 2. Thus, P(A) = 5/7, P( B | A) = 2/3, P( B | A ) = 7/9. 2 5 () 30 P( B | A) P( A) . = 2 53 77 2 = P( A | B ) = P( B | A) P( A) + P( B | A ) P( A ) 3 ( 7 ) + 9 ( 7 ) 44 2.137 Let A = {both balls are white}, and for i = 1, 2, โฆ 5 Ai = both balls selected from bowl i are white. Then โช Ai = A. Bi = bowl i is selected. Then, P( Bi ) = .2 for all i. Chapter 2: Probability 24 Instructorโs Solutions Manual a. P(A) = โ P( Ai | Bi ) P( B i ) = 15 [0 + 25 ( 14 ) + 53 ( 24 ) + 45 ( 43 ) + 1] = 2/5. 3 b. Using Bayesโ rule, P(B3|A) = 502 = 3/20. 50 2.138 Define the events: A: the player wins Bi: a sum of i on first toss Ck: obtain a sum of k before obtaining a 7 12 Now, P( A) = โ P( A โฉ Bi ) . We have that P( A โฉ B2 ) = P( A โฉ B3 ) = P( A โฉ B12 ) = 0. i =1 Also, P( A โฉ B7 ) = P( B7 ) = 366 , P( A โฉ B11 ) = P( B11 ) = 362 . Now, P( A โฉ B4 ) = P(C 4 โฉ B7 ) = P(C4 ) P( B7 ) = 13 ( 363 ) = 363 (using independence Ex. 119). Similarly, P(C5) = P(C9) = 104 , P(C6) = P(C8) = 115 , and P(C10) = 93 . 25 , P( A โฉ B10 ) = 361 . Thus, P( A โฉ B5 ) = P( A โฉ B9 ) = 452 , P( A โฉ B6 ) = P( A โฉ B8 ) = 396 Putting all of this together, P(A) = 0.493. 2.139 From Ex. 1.112, P(Y = 0) = (.02)3 and P(Y = 3) = (.98)3. The event Y = 1 are the events FDF, DFF, and FFD, each having probability (.02)2(.98). So, P(Y = 1) = 3(.02)2(.98). Similarly, P(Y = 2) = 3(.02)2(.98). โ6โ 2.140 The total number of ways to select 3 from 6 refrigerators is โโ โโ = 20. The total number โ 3โ โ 2 โโ 4 โ โโ , y = 0, 1, 2. So, of ways to select y defectives and 3 โ y nondefectives is โโ โโโโ โ y โ โ 3 โ y โ โ 2 โโ 4 โ โโ โโโโ โโ 0 3 P(Y = 0) = โ โ โ โ = 4/20, P(Y = 1) = 4/20, and P(Y = 2) = 12/20. 20 2.141 The events Y = 2, Y = 3, and Y = 4 were found in Ex. 2.120 to have probabilities 1/15, 2/15, and 3/15 (respectively). The event Y = 5 can occur in four ways: DGGGD GDGGD GGDGD GGGDD Each of these possibilities has probability 1/15, so that P(Y = 5) = 4/15. By the complement rule, P(Y = 6) = 5/15. 2.142 Each position has probability 1/4, so every ordering of two positions (from two spins) has โ4โ 1 probability 1/16. The values for Y are 2, 3. P(Y = 2) = โโ โโ = 3/4. So, P(Y = 3) = 1/4. โ 2 โ 16 Chapter 2: Probability 25 Instructorโs Solutions Manual 2.143 Since P( B ) = P( B โฉ A) + P( B โฉ A ) , 1 = P( B โฉ A) P( B โฉ A ) + = P( A | B ) + P( A | B ) . P( B ) P( B ) 2.144 a. S = {16 possibilities of drawing 0 to 4 of the sample points} โ4โ โ4โ โ4โ โ4โ โ4โ b. โโ โโ + โโ โโ + โโ โโ + โโ โโ + โโ โโ = 1 + 4 + 6 + 4 + 1 = 16 = 2 4. โ 0 โ โ 1 โ โ 2โ โ 3โ โ 4โ c. A โช B = {E1, E2, E3, E4}, A โฉ B = {E2}, A โฉ B = 0/ , A โช B = {E2, E4}. 2.145 All 18 orderings are possible, so the total number of orderings is 18! โ 52 โ โ13 โ 2.146 There are โโ โโ ways to draw 5 cards from the deck. For each suit, there are โโ โโ ways โ5โ โ5โ โ13โ โ 52 โ to select 5 cards. Since there are 4 suits, the probability is 4โโ โโ โโ โโ = 0.00248. โ5โ โ 5โ 2.147 The gambler will have a full house if he is dealt {two kings} or {an ace and a king} (there are 47 cards remaining in the deck, two of which are aces and three are kings). โ 3 โ โ 47 โ โ 3 โโ 2 โ โ 47 โ The probabilities of these two events are โโ โโ โโ โโ and โโ โโโโ โโ โโ โโ , respectively. โ2โ โ 2 โ โ 1 โ โ 1 โ โ 2 โ โ 3โ So, the probability of a full house is โโ โโ โ2โ โ 47 โ โ 3 โโ 2 โ โโ โโ + โโ โโโโ โโ โ 2 โ โ 1 โ โ 1 โ โ 47 โ โโ โโ = 0.0083. โ2โ โ12 โ 2.148 Note that โโ โโ = 495 . P(each supplier has at least one component tested) is given by โ4โ โ 3 โโ 4 โโ 5 โ โ 3 โโ 4 โโ 5 โ โ 3 โโ 4 โโ 5 โ โโ โโโโ โโโโ โโ + โโ โโโโ โโโโ โโ + โโ โโโโ โโโโ โโ โ 2 โ โ 1 โ โ 1 โ โ 1 โ โ 2 โ โ 1 โ โ 1 โ โ 1 โ โ 2 โ = 270/475 = 0.545. 495 2.149 Let A be the event that the person has symptom A and define B similarly. Then a. P( A โช B ) = P( A โฉ B ) = 0.4 b. P( A โช B ) = 1 โ P( A โช B ) = 0.6. c. P( A โฉ B | B ) = P( A โฉ B ) / P( B ) = .1/.4 = 0.25 2.150 P(Y = 0) = 0.4, P(Y = 1) = 0.2 + 0.3 = 0.5, P(Y = 2) = 0.1. 2.151 The probability that team A wins in 5 games is p4(1 โ p) and the probability that team B wins in 5 games is p(1 โ p)4. Since there are 4 ways that each team can win in 5 games, the probability is 4[p4(1 โ p) + p(1 โ p)4]. Chapter 2: Probability 26 Instructorโs Solutions Manual 2.152 Let R denote the event that the specimen turns red and N denote the event that the specimen contains nitrates. a. P( R ) = P( R | N ) P( N ) + P( R | N ) P( N ) = .95(.3) + .1(.7) = 0.355. b. Using Bayesโ rule, P(N|R) = .95(.3)/.355 = 0.803. 2.153 Using Bayesโ rule, P( I 1 | H ) = P( H | I 1 ) P( I 1 ) = 0.313. P( H | I 1 ) P( I 1 ) + P( H | I 2 ) P( I 2 ) + P( H | I 3 ) P( I 3 ) 2.154 Let Y = the number of pairs chosen. Then, the possible values are 0, 1, and 2. โ10 โ โ 5โ a. There are โโ โโ = 210 ways to choose 4 socks from 10 and there are โโ โโ 24 = 80 ways โ4โ โ4โ to pick 4 non-matching socks. So, P(Y = 0) = 80/210. โnโ b. Generalizing the above, the probability is โโ โโ 2 2 r โ 2r โ 2.155 a. P(A) = .25 + .1 + .05 + .1 = .5 b. P(AโฉB) = .1 + .05 = 0.15. c. 0.10 d. Using the result from Ex. 2.80, 2.156 a. i. 1 โ 5686/97900 = 0.942 ii. 10560/14113 = 0.748 โ 2n โ โโ โโ . โ 2r โ .25 + .25 โ .15 = 0.875. .4 ii. (97900 โ 43354)/97900 = 0.557 iv. (646+375+568)/11533 = 0.138 b. If the US population in 2002 was known, this could be used to divide into the total number of deaths in 2002 to give a probability. 2.157 Let D denote death due to lung cancer and S denote being a smoker. Thus: P( D ) = P( D | S ) P( S ) + P( D | S ) P( S ) = 10 P( D | S )(.2) + P( D | S )(.8) = 0.006. Thus, P( D | S ) = 0.021 . 2.158 Let W denote the even that the first ball is white and B denote the event that the second ball is black. Then: b (w) w P( B | W ) P(W ) P(W | B ) = = b ww+b+ n w+b b+ n b = P( B | W ) P(W ) + P( B | W ) P(W ) w+b+n w +b + n ( w +b ) + w +b + n ( w +b ) 2.159 Note that S = S โช 0/ , and S and 0/ are disjoint. So, 1 = P(S) = P(S) + P( 0/ ). So, P( 0/ ) = 0. Chapter 2: Probability 27 Instructorโs Solutions Manual 2.160 There are 10 nondefective and 2 defective tubes that have been drawn from the machine, โ12 โ and number of distinct arrangements is โโ โโ = 66. โ2โ a. The probability of observing the specific arrangement is 1/66. b. There are two such arrangements that consist of โruns.โ In addition to what was given in part a, the other is DDNNNNNNNNNNNN. Thus, the probability of two runs is 2/66 = 1/33. 2.161 We must find P(R โค 3) = P(R = 3) + P(R = 2), since the minimum value for R is 2. Id the two Dโs occurs on consecutive trials (but not in positions 1 and 2 or 11 and 12), there are 9 such arrangements. The only other case is a defective in position 1 and 12, so that (combining with Ex. 2.160 with R = 2), there are 12 possibilities. So, P(R โค 3) = 12/66. 2.162 There are 9! ways for the attendant to park the cars. There are 3! ways to park the expensive cars together and there are 7 ways the expensive cars can be next to each other in the 9 spaces. So, the probability is 7(3!)/9! = 1/12. 2.163 Let A be the event that current flows in design A and let B be defined similarly. Design A will function if (1 or 2) & (3 or 4) operate. Design B will function if (1 & 3) or (2 & 4) operate. Denote the event Ri = {relay i operates properly}, i = 1, 2, 3, 4. So, using independence and the addition rule, P(A) = ( R1 โช R2 ) โฉ ( R3 โช R4 ) = (.9 + .9 โ .92)(.9 + .9 โ .92) = 0.9801. P(B) = ( R1 โฉ R3 ) โช ( R2 โฉ R4 ) = .92 + .92 โ (.92)2 = .9639. So, design A has the higher probability. 2.164 Using the notation from Ex. 2.163, P( R1 โฉ R4 | A) = P( R1 โฉ R4 โฉ A) / P( A) . Note that R1 โฉ R4 โฉ A = R1 โฉ R4 , since the event R1 โฉ R4 represents a path for the current to flow. The probability of this above event is .92 = .81, and the conditional probability is in question is .81/.9801 = 0.8264. 2.165 Using the notation from Ex. 2.163, P( R1 โฉ R4 | B ) = P( R1 โฉ R4 โฉ B ) / P( B ) . R1 โฉ R4 โฉ B = ( R1 โฉ R4 ) โฉ ( R1 โฉ R3 ) โช ( R2 โฉ R4 ) = ( R1 โฉ R4 โฉ R3 ) โช ( R2 โฉ R4 ) . The probability of the above event is .93 + .92 – .94 = 0.8829. So, the conditional probability in question is .8829/.9639 = 0.916. โ8โ 2.166 There are โโ โโ = 70 ways to choose the tires. If the best tire the customer has is ranked โ4โ โ 5โ #3, the other three tires are from ranks 4, 5, 6, 7, 8. There are โโ โโ = 10 ways to select โ 3โ three tires from these five, so that the probability is 10/70 = 1/7. Chapter 2: Probability 28 Instructorโs Solutions Manual โ7โ 2.167 If Y = 1, the customer chose the best tire. There are โโ โโ = 35 ways to choose the โ 3โ remaining tires, so P(Y = 1) = 35/70 = .5. โ6โ If Y = 2, the customer chose the second best tire. There are โโ โโ = 20 ways to choose the โ 3โ remaining tires, so P(Y = 2) = 20/70 = 2/7. Using the same logic, P(Y = 4) = 4/70 and so P(Y = 5) = 1/70. 2.168 a. The two other tires picked by the customer must have ranks 4, 5, or 6. So, there are โ 3โ โโ โโ = 3 ways to do this. So, the probability is 3/70. โ2โ b. There are four ways the range can be 4: #1 to #5, #2 to #6, #3 to #7, and #4 to #8. Each has probability 3/70 (as found in part a). So, P(R = 4) = 12/70. c. Similar to parts a and b, P(R = 3) = 5/70, P(R = 5) = 18/70, P(R = 6) = 20/70, and P(R = 7) = 15/70. 2.169 a. For each beer drinker, there are 4! = 24 ways to rank the beers. So there are 243 = 13,824 total sample points. b. In order to achieve a combined score of 4 our less, the given beer may receive at most one score of two and the rest being one. Consider brand A. If a beer drinker assigns a one to A there are still 3! = 6 ways to rank the other brands. So, there are 63 ways for brand A to be assigned all ones. Similarly, brand A can be assigned two ones and one two in 3(3!)3 ways. Thus, some beer may earn a total rank less than or equal to four in 4[63 + 3(3!)3] = 3456 ways. So, the probability is 3456/13824 = 0.25. โ7โ 2.170 There are โโ โโ = 35 ways to select three names from seven. If the first name on the list is โ 3โ โ6โ included, the other two names can be picked โโ โโ = 15 ways. So, the probability is 15/35 โ2โ = 3/7. 2.171 It is stated that the probability that Skylab will hit someone is (unconditionally) 1/150, without regard to where that person lives. If one wants to know the probability condition on living in a certain area, it is not possible to determine. 2.172 Only P( A | B + P( A | B ) = 1 is true for any events A and B. Chapter 2: Probability 29 Instructorโs Solutions Manual 2.173 Define the events: D: item is defective C: item goes through inspection Thus P(D) = .1, P(C|D) = .6, and P(C | D ) = .2. Thus, P (C | D ) P ( D ) = .25. P( D | C ) = P (C | D ) P ( D ) + P (C | D ) P ( D ) 2.174 Let A = athlete disqualified previously B = athlete disqualified next term Then, we know P( B | A ) = .15, P( B | A) = .5, P( A) = .3 . To find P(B), use the law of total probability: P(B) = .3(.5) + .7(.15) = 0.255. 2.175 Note that P(A) = P(B) = P(C) = .5. But, P( A โฉ B โฉ C ) = P(HH) = .25 โ (.5)3. So, they are not mutually independent. 2.176 a. P[( A โช B ) โฉ C )] = P[( A โฉ C ) โช ( B โฉ C )] = P( A โฉ C ) + P( B โฉ C ) โ P( A โฉ B โฉ C ) = P( A) P(C ) + P( B ) P(C ) โ P( A) P( B ) P(C ) = [ P( A) + P( B ) โ P( A) P( B )]P(C ) = P( A โฉ B ) P(C ) b. Similar to part a above. 2.177 a. P(no injury in 50 jumps) = (49/50)50 = 0.364. b. P(at least one injury in 50 jumps) = 1 โ P(no injury in 50 jumps) = 0.636. c. P(no injury in n jumps) = (49/50)n โฅ 0.60, so n is at most 25. 2.178 Define the events: E: person is exposed to the flu F: person gets the flu Consider two employees, one of who is inoculated and one not. The probability of interest is the probability that at least one contracts the flu. Consider the complement: P(at least one gets the flu) = 1 โ P(neither employee gets the flu). For the inoculated employee: P( F ) = P( F โฉ E ) + P( F โฉ E ) = .8(.6) + 1(.4) = 0.88. For the non-inoculated employee: P( F ) = P( F โฉ E ) + P( F โฉ E ) = .1(.6) + 1(.4) = 0.46. So, P(at least one gets the flu) = 1 โ .88(.46) = 0.5952 2.179 a. The gamblers break even if each win three times and lose three times. Considering the โ6โ possible sequences of โwinsโ and โlossesโ, there are โโ โโ = 20 possible orderings. Since โ 3โ each has probability ( 12 ) , the probability of breaking even is 20 ( 12 ) = 0.3125. b. In order for this event to occur, the gambler Jones must have $11 at trial 9 and must win on trial 10. So, in the nine remaining trials, seven โwinsโ and two โlossesโ must be โ9โ placed. So, there are โโ โโ = 36 ways to do this. However, this includes cases where โ2โ Jones would win before the 10th trial. Now, Jones can only win the game on an even trial (since he must gain $6). Included in the 36 possibilities, there are three ways Jones could 6 6 Chapter 2: Probability 30 Instructorโs Solutions Manual win on trial 6: WWWWWWWLL, WWWWWWLLW, WWWWWWLWL, and there are six ways Jones could win on trial 8: LWWWWWWWL, WLWWWWWWL, WWLWWWWWL, WWWLWWWWL, WWWWLWWWL, WWWWWLWWL. So, these nine cases must be 10 removed from the 36. So, the probability is 27 ( 12 ) . 2.180 a. If the patrolman starts in the center of the 16×16 square grid, there are 48 possible paths to take. Only four of these will result in reaching the boundary. Since all possible paths are equally likely, the probability is 4/48 = 1/47. b. Assume the patrolman begins by walking north. There are nine possible paths that will bring him back to the starting point: NNSS, NSNS, NSSN, NESW, NWSE, NWES, NEWS, NSEW, NSWE. By symmetry, there are nine possible paths for each of north, south, east, and west as the starting direction. Thus, there are 36 paths in total that result in returning to the starting point. So, the probability is 36/48 = 9/47. 2.181 We will represent the n balls as 0โs and create the N boxes by placing bars ( | ) between the 0โs. For example if there are 6 balls and 4 boxes, the arrangement 0|00||000 represents one ball in box 1, two balls in box 2, no balls in box 3, and three balls in box 4. Note that six 0โs were need but only 3 bars. In general, n 0โs and N โ 1 bars are needed to โ N + n โ 1โ โโ represent each possible placement of n balls in N boxes. Thus, there are โโ โ N โ1 โ ways to arrange the 0โs and bars. Now, if no two bars are placed next to each other, no box will be empty. So, the N โ 1 bars must be placed in the n โ 1 spaces between the 0โs. โ n โ1 โ โโ , so that the probability is as given in the The total number of ways to do this is โโ โ N โ 1โ problem.

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### Mathematical Statistics With Applications, 7th Edition Solution Manual

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