Mont4e_sm_ch02_sec05 - (c P A | B = P A'∩ B 9 100 = = 0.333(18 9 100 P B 2-82.a P(gas leak =(55 32/107 = 0.813 b P(electric failure|gas leak

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Unformatted text preview: (c) P ( A' | B ' ) = P ( A'∩ B' ) 9 / 100 = = 0.333 (18 + 9) / 100 P( B' ) 2-82.a) P(gas leak) = (55 + 32)/107 = 0.813 b) P(electric failure|gas leak) = (55/107)/(87/102) = 0.632 c) P(gas leak| electric failure) = (55/107)/(72/107) = 0.764 2-83. a) 20/100 b) 19/99 c) (20/100)(19/99) = 0.038 d) If the chips are replaced, the probability would be (20/100) = 0.2 a) 4/499 = 0.0080 b) (5/500)(4/499) = 0.000080 c) (495/500)(494/499) = 0.98 d) 3/498 = 0.0060 e) 4/498 = 0.0080 f) ⎛ 5 ⎞ ⎛ 4 ⎞ ⎛ 3 ⎞ = 4.82x10 −7 ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ 500 ⎠ ⎝ 499 ⎠ ⎝ 498 ⎠ 2-84. 2-85. a)P=(8-1)/(350-1)=0.020 b)P=(8/350) × [(8-1)/(350-1)]=0.000458 c)P=(342/350) × [(342-1)/(350-1)]=0.9547 2-86. 1 36 7 1 (b) 5(36 6 ) 1 (c) 5(365 )5 (a) No, if B ⊂ A , then P(A/B) = P( A ∩ B) P(B) = =1 P(B) P(B) 2-87. A B 2-88. A C B Section 2-5 2-16 2-89. a) P( A ∩ B) = P( A B)P(B) = ( 0.4)( 0.5) = 0.20 b) P( A ′ ∩ B) = P( A ′ B)P(B) = (0.6)(0.5) = 0.30 2-90. P( A ) = P( A ∩ B) + P( A ∩ B ′) = P( A B)P(B) + P( A B ′)P(B ′) = (0.2)(0.8) + (0.3)(0.2) = 0.16 + 0.06 = 0.22 2-91. Let F denote the event that a connector fails. Let W denote the event that a connector is wet. P(F ) = P(F W )P( W ) + P(F W ′)P( W ′) = (0.05)(0.10) + (0.01)(0.90) = 0.014 2-92. Let F denote the event that a roll contains a flaw. Let C denote the event that a roll is cotton. P ( F) = P ( F C ) P ( C ) + P ( F C ′ ) P ( C ′ ) = ( 0. 02 )( 0. 70) + ( 0. 03)( 0. 30) = 0. 023 2-93. Let R denote the event that a product exhibits surface roughness. Let N,A, and W denote the events that the blades are new, average, and worn, respectively. Then, P(R)= P(R|N)P(N) + P(R|A)P(A) + P(R|W)P(W) = (0.01)(0.25) + (0.03) (0.60) + (0.05)(0.15) = 0.028 2-94. Let A denote the event that a respondent is a college graduate and let B denote the event that a voter votes for Bush. P(B)=P(A)P(B|A)+ P(A’)P(B|A’)=0.38 × 0.53+0.62 × 0.5=0.5114 2-95.a) (0.88)(0.27) = 0.2376 b) (0.12)(0.13+0.52) = 0.0.078 2-96. 2-97. a)P=0.13 × 0.73=0.0949 b)P=0.87 × (0.27+0.17)=0.3828 Let A denote a event that the first part selected has excessive shrinkage. Let B denote the event that the second part selected has excessive shrinkage. a) P(B)= P( B A )P(A) + P( B A ')P(A') = (4/24)(5/25) + (5/24)(20/25) = 0.20 b) Let C denote the event that the third part selected has excessive shrinkage. P(C ) = P(C A ∩ B) P( A ∩ B) + P(C A ∩ B' ) P( A ∩ B' ) + P(C A'∩ B) P( A'∩ B) + P(C A'∩ B' ) P( A'∩ B' ) 3 ⎛ 4 ⎞⎛ 5 ⎞ 4 ⎛ 20 ⎞⎛ 5 ⎞ 4 ⎛ 5 ⎞⎛ 20 ⎞ 5 ⎛ 19 ⎞⎛ 20 ⎞ ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ 23 ⎝ 24 ⎠⎝ 25 ⎠ 23 ⎝ 24 ⎠⎝ 25 ⎠ 23 ⎝ 24 ⎠⎝ 25 ⎠ 23 ⎝ 24 ⎠⎝ 25 ⎠ = 0.20 = 2-98. Let A and B denote the events that the first and second chips selected are defective, respectively. 2-17 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 ...
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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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