plugin-HW_02_Solutions

# plugin-HW_02_Solutions - HW#2 Solutions(Fall 2011 Assigned...

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1 HW #2 Solutions (Fall 2011) Assigned problems : 2.24, 2.37, 2.48, 2.64, 2.65 2.24 (a) Let A, B denote the event of the respective engineers spotting the error. Let E denote the event that the error is spotted, P(E) = P(A B) = P(A) + P(B) – P(AB) = P(A) + P(B) – P(A)P(B) = 0.8 + 0.9 – 0.8 × 0.9 = 0.98 (b) “Spotted by A alone” implies that B failed to spot it, hence the required probability is P(AB’|E) = P(AB’ E)/P(E) = P(E|AB’)P(AB’) / P(E) = 1 × P(A)P(B’) / P(E) = 0.8 × 0.1 / 0.98 0.082 (c) With these 3 engineers checking it, the probability of not finding the error is P(C 1 ’C 2 ’C 3 ’) = P(C 1 ’)P(C 2 ’)P(C 3 ’) = (1 – 0.75) 3 , hence P(error spotted) = 1 – (1 – 0.75) 3 0.984, which is higher than the 0.98 in (a), so the team of 3 is better. (d) The probability that the first error is detected (event D 1 ) has been calculated in (a) to be 0.98. However, since statistical independence is given, detection of the second error (event D 2 ) still has the same probability. Hence P(D 1 D 2 ) = P(D 1 )P(D 2 ) = 0.98 2 0.960

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2 2.37 (a) Let subscripts 1 and 2 denote “after first earthquake” and “after second earthquake”. Note that (i)
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plugin-HW_02_Solutions - HW#2 Solutions(Fall 2011 Assigned...

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