Homework#6_Solutions

Homework#6_Solutions - ENGINEERING PROBABILITY Spring 2012...

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Unformatted text preview: ENGINEERING PROBABILITY Spring 2012 Homework #6 _ Solutions Topic More on Bayes Rule and Independence 2155. 2157. Let A = excellent surface finish; B = excellent length a) P(A) = 82/100 = 0.82 b) P(B) = 90/100 = 0.90 c) P(A') = 1 0.82 = 0.18 d) P(AB) = 80/100 = 0.80 e) P(AB) = 0.92 f) P(A'B) = 0.98 Let B denote the event that a glass breaks. Let L denote the event that large packaging is used. P(B)= P(B|L)P(L) + P(B|L')P(L') = 0.01(0.60) + 0.02(0.40) = 0.014 Assignment Due Next Class 2-155, 157, 160, 163, 170, 176, 178 2160. a) 345/357 b) 5/13 2163. a) P(A) = 0.03 b) P(A') = 0.97 c) P(B|A) = 0.40 d) P(B|A') = 0.05 e) P( A B ) = P( B A )P(A) = (0.40)(0.03) = 0.012 f) P( A B ') = P( B' A )P(A) = (0.60)(0.03) = 0.018 g) P(B) = P( B A )P(A) + P( B A ')P(A') = (0.40)(0.03) + (0.05)(0.97) = 0.0605 1 2170. a) 0.25 b) 0.75 2176. a) (0.20)(0.30) +(0.7)(0.9) = 0.69 2178. Let A,B denote the event that the first, second portion of the circuit operates. Then, P(A) = (0.99)(0.99)(0.9) = 0.998 P(B) = 0.9+0.9(0.9)(0.9) = 0.99 and P( A B ) = P(A) P(B) = (0.998) (0.99) = 0.988 (0.99)(0.99)+0.9 2 ...
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