mont4e_sm_ch02_sec04

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online