CEE 202 Homework 3 Solutions

# 80254 answer 08025 you can also

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Unformatted text preview: ∩~W) = 0.04 * (1 ­0.07) = 0.0372; P(~M ∩~W) = (1 ­0.04) * (1 ­ 0.07) = 0.8928 Total probability: P(D) = P(D | M ∩~W) P(M ∩~W) + P(D |~M ∩W) P(~M ∩W) + P(D |M ∩W) P(M ∩W) + P(D |~M ∩ ~W) P(~M ∩ ~W) = 0.1 * 0.0372 + 0.2* 0.0672 + 0.6* 0.0028 + 0 * 0.8928 = 0.01884 ANSWER: P(D)= 0.0188 Some students had a wrong solution but because of rounding off the answer was numerically almost correct. However, the method was incorrect. An example is: P(D) = P(D | M ∩~W) P(M) + P(D |~M ∩W) P(W) That is they have forgotten to account for: P(D |M ∩W) P(M ∩W). [ P(D |~M ∩ ~W) P(~M ∩ ~W) does not matter as P(D |~M ∩ ~W) = 0] 4/9 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) Homework 3 Bayes’ Theorem and Probability Dist. b) If a defective headset is discovered, what is the probability that careless workers had something to do with the problem? There is only one combination of events that results in a defect NOT being caused by workers: (M ∩ ~W) Bayes’ theorem for event NOT being caused by workers P(M ∩ ~W | D) = P(D | M ∩ ~W) P(M ∩ ~W) / P(D) = 0.1 * 0.0372 / 0.01884 = 0.19745 The event of the defect being CAUSED by workers is the complement: P(workersCaused) = 1 – P(~workersCaused) = 0.80254 ANSWER: 0.8025 You can also do this problem by adding the probabilities of the two events in which workers played a role and dividing by...
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## This document was uploaded on 03/06/2014 for the course CEE 202 at University of Illinois, Urbana Champaign.

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