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20 when both causes occur the probability is 06 define

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Unformatted text preview: aused by overloaded truck (or trucks)? Let’s find the probability that the damage was caused by normal trucks. = P(NN | D) = P(D ∩ NN) / P(D) = P(D | NN) * P(NN) / P(D) {Bayes’ theorem—you could have written this first} = P(D | NN) * P(N) * P(N)} / P(D) = 0.001 * (0.9*0.9) /0.013 = 0.063 P(overload|D) = 1 ­P(normal|D) = 0.937 3/9 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) Homework 3 Bayes’ Theorem and Probability Dist. 4) Defects in the headsets produced by HearYourselfThink, Inc. can be the result of one of the following independent causes:  ­ Malfunction of the machinery, which occurs 4% of the production time  ­ Carelessness of workers, which occurs 7% of the production time There are no other causes of defective articles. If only the machinery malfunctions, the probability of a defective headset is 0.10. If only the workers are careless, the probability of a defective headset is 0.20. When both causes occur, the probability is 0.6. Define: P(D) = probability of defect Given: P(M) = probability that machinery malfunctions = 0.04 P(W) = probability that workers are careless = 0.07 P(D | M ∩~W) = 0.10 P(D | ~M ∩ W) = 0.20 P(D | M ∩ W) = 0.6 a) What is the probability of producing a defective headset? Want: P(D) The following events are mutually exclusive and collectively exhaustive: (M ∩~W) (~M ∩W) (M ∩ W) (~M ∩~W) Since M and W are statistically independent, P(M ∩W) = P(M) P(W) = 0.04 * 0.07 = 0.0028 Likewise, P(~M ∩W) = (1 ­0.04) * 0.07 = 0.0672; P(M...
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