Unformatted text preview: Homework #4 Solutions Topic Total probability rule and independence 2109. 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, 2111. a) (0.88)(0.27) = 0.2376 b) (0.12)(0.13+0.52) = 0.0.078 2122. 2123. 2124. 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. 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. P(R)= P(RN)P(N) + P(RA)P(A) + P(RW)P(W) = (0.01)(0.25) + (0.03) (0.60) + (0.05)(0.15) = 0.028 Assignment Due Next Class 2109, 111, 122,123,124, 127, 131, 136 2127. 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 independent. 2131. (a) Let I and G denote an infested and good sample. There are 3 ways to obtain four consecutive samples showing the signs of the infestation: IIIIGG, GIIIIG, GGIIII. Therefore, the probability is 3 (0.2 4 0 .8 2 ) 0 .003072 b) P(BA) = P(A B)/P(A) = (22/100)/(30/100) = 0.733 not 1 (b) There are 10 ways to obtain three out of four consecutive samples showing the signs of infestation. The probability is 10 (0.2 3 * 0.8 3 ) 0.04096 2136. 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(AB) = (0.504)(0.8574) = 0.4321 Therefore, the probability that the circuit operates = P(AB) = P(A) +P(B) P(AB) = 0.9293 2 ...
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 Spring '12
 SUSANALBIN
 3rd millennium, 22nd century, total probability rule, devices function, consecutive samples

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