mont4e_sm_ch02_sec06

# mont4e_sm_ch02_sec06 - a) P(B) = P(B|A)P(A) + P(B|A')P(A')...

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Unformatted text preview: a) P(B) = P(B|A)P(A) + P(B|A')P(A') = (19/99)(20/100) + (20/99)(80/100) = 0.2 b) Let C denote the event that the third chip selected is defective. P( A ∩ B ∩ C ) = P(C A ∩ B) P( A ∩ B) = P(C A ∩ B) P( B A) P( A) 18 ⎛ 19 ⎞⎛ 20 ⎞ ⎜ ⎟⎜ ⎟ 98 ⎝ 99 ⎠⎝ 100 ⎠ = 0.00705 = 2-99. Open surgery success 192 81 273 failure 71 6 77 sample size 263 87 350 sample percentage 0.751428571 0.248571429 1 conditional success rate 0.73% 0.93% 0.78 large stone small stone overall summary PN large stone small stone overall summary success 55 234 289 failure 25 36 61 sample size 80 270 350 sample percentage 0.228571429 0.771428571 1 conditional success rate 69% 83% 0.83 The reason is that the overall success rate is dependent on both the success rates conditioned on the two groups and the probability of the groups. It is the weighted average of the group success rate weighted by the group size; instead of the direct average. P(overall success)=P(large stone)P(success| large stone)+ P(small stone)P(success| small stone). For open surgery, the dominant group (large stone) has a smaller success rate while for PN, the dominant group (small stone) has a larger success rate. Section 2-6 2-100. 2-101. Because P( A B ) ≠ P(A), the events are not independent. P(A') = 1 - P(A) = 0.7 and P( A ' B ) = 1 - P( A B ) = 0.7 Therefore, A' and B are independent events. 2-102. If A and B are mutually exclusive, then P( A ∩ B ) = 0 and P(A)P(B) = 0.04. Therefore, A and B are not independent. a) P( B A ) = 4/499 and P( B) = P( B A) P( A) + P( B A' ) P( A' ) = (4 / 499)(5 / 500) + (5 / 499)(495 / 500) = 5 / 500 2-103. Therefore, A and B are not independent. b) A and B are independent. 2-104. P( A ∩ B ) = 70/100, P(A) = 86/100, P(B) = 77/100. Then, P( A ∩ B ) ≠ P(A)P(B), so A and B are not independent. a) P( A ∩ B )= 22/100, P(A) = 30/100, P(B) = 77/100, Then P( A ∩ B ) ≠ P(A)P(B), therefore, A and B are not independent. b) P(B|A) = P(A ∩ B)/P(A) = (22/100)/(30/100) = 0.733 2-105. 2-18 2-106. (a) P = (0.001) = 10 2 2 −6 (b) P = 1 − (0.999) = 0.002 2-107. It is useful to work one of these exercises with care to illustrate the laws of probability. Let Hi denote the event that the ith sample contains high levels of contamination. ' ' ' ' ' ' ' ' ' ' a) P(H1 ∩ H2 ∩ H3 ∩ H4 ∩ H5 ) = P(H1)P(H2 )P(H3 )P(H4 )P(H5 ) by independence. Also, P(Hi' ) = 0.9 . Therefore, the answer is 0.9 5 = 0.59 ' ' b) A1 = (H1 ∩ H'2 ∩ H'3 ∩ H4 ∩ H5 ) ' A 2 = (H1 ∩ H2 ∩ H'3 ∩ H'4 ∩ H'5 ) ' ' ' A 3 = (H1 ∩ H2 ∩ H3 ∩ H4 ∩ H'5 ) ' ' ' A 4 = (H1 ∩ H2 ∩ H3 ∩ H4 ∩ H'5 ) ' ' ' ' A 5 = (H1 ∩ H2 ∩ H3 ∩ H4 ∩ H5 ) The requested probability is the probability of the union A 1 ∪ A 2 ∪ A 3 ∪ A 4 ∪ A 5 and these events are mutually exclusive. Also, by independence P( A i ) = 0.9 4 (0.1) = 0.0656 . Therefore, the answer is 5(0.0656) = 0.328. c) Let B denote the event that no sample contains high levels of contamination. The requested probability is P(B') = 1 - P(B). From part (a), P(B') = 1 - 0.59 = 0.41. 2-108. Let A i denote the event that the ith bit is a one. a) By independence P( A 1 ∩ A 2 ∩...∩ A 10 ) = P( A 1 )P( A 2 )...P( A 10 ) = ( 1 )10 = 0.000976 2 ' ' ' c b) By independence, P ( A 1 ∩ A '2 ∩... ∩ A 10 ) = P ( A 1 ) P ( A '2 )... P ( A 10 ) = ( 1 ) 10 = 0. 000976 2 c) The probability of the following sequence is 1 ' ' ' ' ' P( A 1 ∩ A 2 ∩ A 3 ∩ A 4 ∩ A 5 ∩ A 6 ∩ A 7 ∩ A 8 ∩ A 9 ∩ A10 ) = ( )10 , by independence. The number of 2 10 10 ! sequences consisting of five "1"'s, and five "0"'s is 10 = = 252 . The answer is 252⎛ 1 ⎞ = 0.246 ⎜⎟ 5 5! 5! ⎝ 2⎠ () 2-109. (a) 3(0.2 ) =0.0048 (b) 3(4 * 0.2 * 0.8) =0.0768 3 4 2-110. (a) P = (0.8) = 0.4096 4 (b) P = 1 − 0.2 − 0.8 × 0.2 = 0.64 (c) Probability defeats all four in a game = 0.84 = 0.4096. Probability defeats all four at least once = 1 – (1 – 0.4096)3 = 0.7942 2-111. (a) The probability that one technician obtains equivalence at 100 mL is 0.1. So the probability that both technicians obtain equivalence at 100 mL is 0.1 = 0.01 . (b) The probability that one technician obtains equivalence between 98 and 104 mL is 0.7. So the probability that both technicians obtain equivalence between 98 and 104 mL is 2 0.7 2 = 0.49 . (c) The probability that the average volume at equivalence from the technician is 100 mL is 9(0.12 ) = 0.09 . 2-19 2-112. 10 6 = 10 −10 16 10 1 (b) P = 0.25 × ( ) = 0.020833 12 (a) P = Let A denote the event that a sample is produced in cavity one of the mold. 1 a) By independence, P( A1 ∩ A 2 ∩ A 3 ∩ A 4 ∩ A 5 ) = ( )5 = 0.00003 8 b) Let Bi be the event that all five samples are produced in cavity i. Because the B's are mutually exclusive, P(B1 ∪ B 2 ∪...∪B 8 ) = P(B1) + P(B 2 )+...+P(B 8 ) 2-113. 2-114. 1 1 From part a., P(Bi ) = ( )5 . Therefore, the answer is 8( )5 = 0.00024 8 8 147 ' c) By independence, P( A 1 ∩ A 2 ∩ A 3 ∩ A 4 ∩ A 5 ) = ( ) ( ) . The number of sequences in 8 8 1 7 which four out of five samples are from cavity one is 5. Therefore, the answer is 5( ) 4 ( ) = 0.00107 . 8 8 Let A denote the upper devices function. Let B denote the lower devices function. P(A) = (0.9)(0.8)(0.7) = 0.504 P(B) = (0.95)(0.95)(0.95) = 0.8574 P(A∩B) = (0.504)(0.8574) = 0.4321 Therefore, the probability that the circuit operates = P(A∪B) = P(A) +P(B) − P(A∩B) = 0.9293 [1-(0.1)(0.05)][1-(0.1)(0.05)][1-(0.2)(0.1)] = 0.9702 2-115. Section 2-7 2-116. Because, P( A B ) P(B) = P( A ∩ B ) = P( B A ) P(A), P( B A) = P( A B) P( B) P( A) P( A B) P( B) P( A) = 0.7(0.2) = 0.28 0.5 P( A B) P( B ) P ( A B) P( B) + P( A B ' ) P( B ' ) P( B A) = 2-117. = = 2-118. 0.4 × 0.8 = 0.89 0.4 × 0.8 + 0.2 × 0.2 Let F denote a fraudulent user and let T denote a user that originates calls from two or more metropolitan areas in a day. Then, P(T F ) P( F ) 0.30(0.0001) = = 0.003 P( F T ) = P(T F ) P( F ) + P(T F ' ) P( F ' ) 0.30(0.0001) + 0.01(.9999) 2-119. (a) P=(0.31)(0.978)+(0.27)(0.981)+(0.21)(0.965)+(0.13)(0.992)+(0.08)(0.959) = 0.97638 (b) P = (0.21)(0.965) = 0.207552 0.97638 2-120. Let A denote the event that a respondent is a college graduate and let B denote the event that a voter votes for Bush. 2-20 ...
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## This note was uploaded on 02/17/2010 for the course IE 33214 taught by Professor Feng during the Spring '10 term at University of Illinois at Urbana–Champaign.

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