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Assignment 1: Fundamentals Concepts and Failure Distribution [EN 9116] 1. A piece of equipment contains six identical items and it is known that three of them are defective. The items are tested one after the other until the three defective items are found. a) What is the probability that the testing process is stopped on the (i) third test (ii) fourth test. b) If the process is stopped on the fourth test, what is the probability that the first item is not defective. Solution: Let D and N denote the events of having a defective and non-defective item respectively. a) (i) For the testing process to be stopped in 3 rd test, defective items should be found in 1 st , 2 nd and 3 rd tests, thus the sample space is DDD only P(testing process is stopped on the 3 rd test)= P(DDD) = 3/6 x 2/5 x 1/4 = 0.05 (ii) For the testing process to be stopped in 4 th test, 3 rd defective item should be found in 4 th test and 1 st and 2 nd defective items should be found in 1 st , 2 nd or 3 rd test. This is represented by the sample space: NDDD + DNDD + DDND P(testing process is stopped in the 4 th test) = P(NDDD) + P(DNDD) + P(DDND) = 3/6 x 3/5 x 2/4 x 1/3 + 3/6 x 3/5 x 2/4 x 1/3 + 3/6 x 2/5 x 3/4 x 1/3 = 0.05 + 0.05 + 0.05 = 0.15 b) P(1 st item is not defective|testing process is stopped on the 4 th test) = P(NDDD) / 0.15 = 0.05 / 0.15 = 0.33
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Assignment 1: Fundamentals Concepts and Failure Distribution [EN 9116] 2.
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