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# Solucionario Walpole 8 ED

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system works. P (A\u2032 | S \u2032) = P (A\u2032\u2229S\u2032) P (S\u2032) = P (A \u2032)(1\u2212P (C\u2229D\u2229E)) 1\u2212P (S) = (0.3)[1\u2212(0.8)(0.8)(0.8)] 1\u22120.75112 = 0.588. 2.101 Consider the events: C: an adult selected has cancer, D: the adult is diagnosed as having cancer. P (C) = 0.05, P (D | C) = 0.78, P (C \u2032) = 0.95 and P (D | C \u2032) = 0.06. So, P (D) = P (C \u2229D) + P (C \u2032 \u2229D) = (0.05)(0.78) + (0.95)(0.06) = 0.096. 24 Chapter 2 Probability 2.102 Let S1, S2, S3, and S4 represent the events that a person is speeding as he passes through the respective locations and let R represent the event that the radar traps is operating resulting in a speeding ticket. Then the probability that he receives a speeding ticket: P (R) = 4\u2211 i=1 P (R | Si)P (Si) = (0.4)(0.2) + (0.3)(0.1) + (0.2)(0.5) + (0.3)(0.2) = 0.27. 2.103 P (C | D) = P (C\u2229D) P (D) = 0.039 0.096 = 0.40625. 2.104 P (S2 | R) = P (R\u2229 S2)P (R) = 0.030.27 = 1/9. 2.105 Consider the events: A: no expiration date, B1: John is the inspector, P (B1) = 0.20 and P (A | B1) = 0.005, B2: Tom is the inspector, P (B2) = 0.60 and P (A | B2) = 0.010, B3: Jeff is the inspector, P (B3) = 0.15 and P (A | B3) = 0.011, B4: Pat is the inspector, P (B4) = 0.05 and P (A | B4) = 0.005, P (B1 | A) = (0.005)(0.20)(0.005)(0.20)+(0.010)(0.60)+(0.011)(0.15)+(0.005)(0.05) = 0.1124. 2.106 Consider the events E: a malfunction by other human errors, A: station A, B: station B, and C: station C. P (C |E) = P (E | C)P (C) P (E | A)P (A)+P (E | B)P (B)+P (E | C)P (C) = (5/10)(10/43) (7/18)(18/43)+(7/15)(15/43)+(5/10)(10/43) = 0.1163 0.4419 = 0.2632. 2.107 (a) P (A \u2229 B \u2229 C) = P (C | A \u2229B)P (B | A)P (A) = (0.20)(0.75)(0.3) = 0.045. (b) P (B\u2032 \u2229 C) = P (A \u2229B\u2032 \u2229 C) + P (A\u2032 \u2229B\u2032 \u2229 C) = P (C | A \u2229B\u2032)P (B\u2032 | A)P (A) + P (C | A\u2032\u2229B\u2032)P (B\u2032 | A\u2032)P (A\u2032) = (0.80)(1\u22120.75)(0.3)+(0.90)(1\u22120.20)(1\u22120.3) = 0.564. (c) Use similar argument as in (a) and (b), P (C) = P (A\u2229B \u2229C)+P (A\u2229B\u2032 \u2229C)+ P (A\u2032 \u2229 B \u2229 C) + P (A\u2032 \u2229B\u2032 \u2229 C) = 0.045 + 0.060 + 0.021 + 0.504 = 0.630. (d) P (A | B\u2032 \u2229 C) = P (A \u2229B\u2032 \u2229 C)/P (B\u2032 \u2229 C) = (0.06)(0.564) = 0.1064. 2.108 Consider the events: A: a customer purchases latex paint, A\u2032: a customer purchases semigloss paint, B: a customer purchases rollers. P (A | B) = P (B | A)P (A) P (B | A)P (A)+P (B | A\u2032)P (A\u2032) = (0.60)(0.75) (0.60)(0.75)+(0.25)(0.30) = 0.857. 2.109 Consider the events: G: guilty of committing a crime, I: innocent of the crime, i: judged innocent of the crime, g: judged guilty of the crime. P (I | g) = P (g | I)P (I) P (g | G)P (G)+P (g | I)P (I) = (0.01)(0.95) (0.05)(0.90)+(0.01)(0.95) = 0.1743. Solutions for Exercises in Chapter 2 25 2.110 Let Ai be the event that the ith patient is allergic to some type of week. (a) P (A1 \u2229 A2 \u2229 A3 \u2229 A\u20324) + P (A1 \u2229 A2 \u2229 A\u20323 \u2229 A4) + P (A1 \u2229 A\u20322 \u2229 A3 \u2229 A4) + P (A \u2032 1 \u2229 A2 \u2229 A3 \u2229 A4) = P (A1)P (A2)P (A3)P (A\u20324) + P (A1)P (A2)P (A\u20323)P (A4) + P (A1)P (A \u2032 2)P (A3)P (A4) + P (A \u2032 1)P (A2)P (A3)P (A4) = (4)(1/2) 4 = 1/4. (b) P (A \u2032 1 \u2229A\u20322 \u2229A\u20323 \u2229 A\u20324) = P (A\u20321)P (A\u20322)P (A\u20323)P (A\u20324) = (1/2)4 = 1/16. 2.111 No solution necessary. 2.112 (a) 0.28 + 0.10 + 0.17 = 0.55. (b) 1\u2212 0.17 = 0.83. (c) 0.10 + 0.17 = 0.27. 2.113 P = (134 )( 13 6 )( 13 1 )( 13 2 ) (5213) . 2.114 (a) P (M1 \u2229M2 \u2229M3 \u2229M4) = (0.1)4 = 0.0001, where Mi represents that ith person make a mistake. (b) P (J \u2229 C \u2229R\u2032 \u2229W \u2032) = (0.1)(0.1)(0.9)(0.9) = 0.0081. 2.115 Let R, S, and L represent the events that a client is assigned a room at the Ramada Inn, Sheraton, and Lakeview Motor Lodge, respectively, and let F represents the event that the plumbing is faulty. (a) P (F ) = P (F | R)P (R) + P (F | S)P (S) + P (F | L)P (L) = (0.05)(0.2) + (0.04)(0.4) + (0.08)(0.3) = 0.054. (b) P (L | F ) = (0.08)(0.3) 0.054 = 4 9 . 2.116 (a) There are ( 9 3 ) = 84 possible committees. (b) There are ( 4 1 )( 5 2 ) = 40 possible committees. (c) There are ( 3 1 )( 1 1 )( 5 1 ) = 15 possible committees. 2.117 Denote by R the event that a patient survives. Then P (R) = 0.8. (a) P (R1 \u2229 R2 \u2229 R\u20323) + P (R1 \u2229 R\u20322 \u2229 R3)P (R\u20321 \u2229 R2 \u2229 R3) = P (R1)P (R2)P (R\u20323) + P (R1)P (R \u2032 2)P (R3) + P (R \u2032 1)P (R2)P (R3) = (3)(0.8)(0.8)(0.2) = 0.384. (b) P (R1 \u2229R2 \u2229 R3) = P (R1)P (R2)P (R3) = (0.8)3 = 0.512. 2.118 Consider events M : an inmate is a male, N : an inmate is under 25 years of age. P (M \u2032 \u2229N \u2032) = P (M \u2032) + P (N \u2032)\u2212 P (M \u2032 \u222aN \u2032) = 2/5 + 1/3\u2212 5/8 = 13/120. 2.119 There are ( 4 3 )( 5 3 )( 6 3 ) = 800 possible selections. 26 Chapter 2 Probability 2.120 Consider the events: Bi: a black ball is drawn on the ith drawl, Gi: a green ball is drawn on the ith drawl. (a) P (B1\u2229B2\u2229B3)+P (G1\u2229G2\u2229G3) = (6/10)(6/10)(6/10)+(4/10)(4/10)(4/10) = 7/25. (b) The probability that each color is represented is 1\u2212 7/25 = 18/25. 2.121 The total number of ways to receive 2 or 3 defective sets among 5 that are purchased is ( 3 2 )( 9 3 ) + ( 3 3 )( 9 2 ) = 288. 2.122 A Venn diagram is shown next. A B C S 1 2 3 4 5 6 7 8 (a) (A \u2229B)\u2032: 1, 2, 3, 6, 7, 8. (b) (A \u222aB)\u2032: 1, 6. (c) (A \u2229 C) \u222a B: 3, 4, 5, 7, 8. 2.123 Consider the events: O: overrun, A: consulting firm A, B: consulting firm B, C: consulting firm C. (a) P (C |O) = P (O | C)P (C) P (O | A)P (A)+P (O | B)P (B)+P (O | C)P (C) = (0.15)(0.25) (0.05)(0.40)+(0.03)(0.35)+(0.15)(0.25) = 0.0375 0.0680 = 0.5515. (b) P (A | O) = (0.05)(0.40) 0.0680 = 0.2941. 2.124 (a) 36; (b) 12; (c) order is not important. 2.125 (a) 1 (362 ) = 0.0016; (b) (121 )( 24 1 ) (362 ) = 288 630 = 0.4571. Solutions for Exercises in Chapter 2 27 2.126 Consider the events: C: a woman over 60 has the cancer, P : the test gives a positive result. So, P (C) = 0.07, P (P \u2032 | C) = 0.1 and P (P | C \u2032) = 0.05. P (C | P \u2032) = P (P \u2032 | C)P (C) P (P \u2032 | C)P (C)+P (P \u2032 | C\u2032)P (C\u2032) = (0.1)(0.07) (0.1)(0.07)+(1\u22120.05)(1\u22120.07) = 0.007 0.8905 = 0.00786. 2.127 Consider the events: A: two nondefective components are selected, N : a lot does not contain defective components, P (N) = 0.6, P (A | N) = 1, O: a lot contains one defective component, P (O) = 0.3, P (A | O) = ( 19 2 ) (202 ) = 9 10 , T : a lot contains two defective components,P (T ) = 0.1, P (A | T ) = ( 18 2 ) (202 ) = 153 190 . (a) P (N | A) = P (A | N)P (N) P (A | N)P (N)+P (A | O)P (O)+P (A | T )P (T ) = (1)(0.6) (1)(0.6)+(9/10)(0.3)+(153/190)(0.1) = 0.6 0.9505 = 0.6312; (b) P (O | A) = (9/10)(0.3) 0.9505 = 0.2841; (c) P (T | A) = 1\u2212 0.6312\u2212 0.2841 = 0.0847. 2.128 Consider events: D: a person has the rare disease, P (D) = 1/500, P : the test shows a positive result, P (P | D) = 0.95 and P (P | D\u2032) = 0.01. P (D | P ) = P (P | D)P (D) P (P | D)P (D)+P (P | D\u2032)P (D\u2032) = (0.95)(1/500) (0.95)(1/500)+(0.01)(1\u22121/500) = 0.1599. 2.129 Consider the events: 1: engineer 1, P (1) = 0.7, and 2: engineer 2, P (2) = 0.3, E: an error has occurred in estimating cost, P (E | 1) = 0.02 and P (E | 2) = 0.04. P (1 | E) = P (E | 1)P (1) P (E | 1)P (1)+P (E | 2)P (2) = (0.02)(0.7) (0.02)(0.7)+(0.04)(0.3) = 0.5385, and P (2 | E) = 1\u2212 0.5385 = 0.4615. So, more likely engineer 1 did the job. 2.130 Consider the events: D: an item is defective (a) P (D1D2D3) = P (D1)P (D2)P (D3) = (0.2) 3 = 0.008. (b) P (three out of four are defectives) = ( 4 3 ) (0.2)3(1\u2212 0.2) = 0.0256. 2.131 LetA be the event that an injured worker is admitted to the hospital andN be the event that an injured worker is back to work the next day. P (A) = 0.10, P (N) = 0.15 and P (A\u2229N) = 0.02. So, P (A\u222aN) = P (A)+P (N)\u2212P (A\u2229N) = 0.1+0.15\u22120.02 = 0.23. 2.132 Consider the events: T : an operator is trained, P (T ) = 0.5, M an operator meets quota, P (M | T ) = 0.9 and P (M | T \u2032) = 0.65. P (T | M) = P (M | T )P (T ) P