mont4e_sm_ch02_sec04

# mont4e_sm_ch02_sec04 - c P A ∩ B = 0 because A ∩ B =...

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Unformatted text preview: c) P( A ∩ B ) = 0 , because A ∩ B = ∅ d) P( ( A ∪ B ) ∩ C ) = 0, because ( A ∪ B ) ∩ C = ( A ∩ C ) ∪ ( B ∩ C ) e) P( A′ ∩ B′ ∩ C ′ ) =1-[ P(A) + P(B) + P(C)] = 1-(0.2+0.3+0.4) = 0.1 =∅ 2-68. (a) P(Caused by sports) = P(Caused by contact sports or by noncontact sports) = P(Caused by contact sports) + P(Caused by noncontact sports) =0.46+0.44 =0.9 (b) 1- P(Caused by sports)=0.1. 2-69.a) 70/100 = 0.70 b) (79+86-70)/100 = 0.95 c) No, P( A ∩ B ) ≠ 0 2-70. (a) P(High temperature and high conductivity)= 74/100 =0.74 (b) P(Low temperature or low conductivity) = P(Low temperature) + P(Low conductivity) – P(Low temperature and low conductivity) =(8+3)/100 + (15+3)/100 – 3/100 =0.26 (c) No, they are not mutually exclusive. Because P(Low temperature) + P(Low conductivity) =(8+3)/100 + (15+3)/100 =0.29, which is not equal to P(Low temperature or low conductivity). a) 350/370 345 + 5 + 12 362 b) = 370 370 345 + 5 + 8 358 c) = 370 370 d) 345/370 2-71. 2-72.a) 170/190 = 17/19 b) 7/190 2-73.a) P(unsatisfactory) = (5+10-2)/130 = 13/130 b) P(both criteria satisfactory) = 117/130 = 0.90, No 2-74. (a) 5/36 (b) 5/36 (c) P ( A ∩ B ) = P( A) P( B) = 0.01929 (d) P ( A ∪ B ) = P( A) + P ( B ) = 0.2585 Section 2-4 2-75. a) P(A) = 86/100 b) P(B) = 79/100 2-14 c) P( A B ) = d) P( B A ) = P ( A ∩ B) 70 / 100 70 = = P( B) 79 / 100 79 P( A ∩ B) 70 / 100 70 = = P( A) 86 / 100 86 2-76. 7 + 32 = 0.39 100 13 + 7 = 0.2 (b) P ( B ) = 100 P( A ∩ B) 7 / 100 = = 0.35 (c) P ( A | B ) = 20 / 100 P( B) P( A ∩ B) 7 / 100 = = 0.1795 (d) P ( B | A) = 39 / 100 P( A) (a) P ( A) = 2-77. Let A denote the event that a leaf completes the color transformation and let B denote the event that a leaf completes the textural transformation. The total number of experiments is 300. P( A ∩ B) 243 / 300 = = 0.903 (243 + 26) / 300 P( A) P( A ∩ B' ) 26 / 300 = = 0.591 (b) P ( A | B ' ) = (18 + 26) / 300 P( B' ) (a) P ( B | A) = 2-78.a) 0.82 b) 0.90 c) 8/9 = 0.889 d) 80/82 = 0.9756 e) 80/82 = 0.9756 f) 2/10 = 0.20 2-79. 2-80. a) 12/100 b) 12/28 c) 34/122 a) P(A) = 0.05 + 0.10 = 0.15 P( A ∩ B ) 0.04 + 0.07 b) P(A|B) = = = 0.153 0.72 P( B) c) P(B) = 0.72 d) P(B|A) = P( A ∩ B) = 0.04 + 0.07 = 0.733 P( A) 0.15 e) P(A ∩ B) = 0.04 +0.07 = 0.11 f) P(A ∪ B) = 0.15 + 0.72 – 0.11 = 0.76 2-81. Let A denote the event that autolysis is high and let B denote the event that putrefaction is high. The total number of experiments is 100. P( A ∩ B' ) 18 / 100 = = 0.5625 (14 + 18) / 100 P( A) P( A ∩ B) 14 / 100 = = 0.1918 (b) P ( A | B ) = (14 + 59) / 100 P( B) (a) P ( B ' | A) = 2-15 (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 ...
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