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Better to find the probability that 2 or fewer pipes

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Unformatted text preview: = number of pipes failing a) You could do a very long sum of probability of 3 or more pipes will fail, but this would take a long time. Better to find the probability that 2 or fewer pipes will fail. Then find out the probability of at least 3 pipes will fail by deducting the probability of 2 or fewer will fail from 1. Here are the calculations. Let ‘p’ be the probability of failure, ‘q’ be the probability of success which is (1 ­p). Then p=0.001, and q =0.999 No. trials(n) 4000 p failure 0.001 q success 0.999 ȹ n ȹ n(x)= ȹ ȹ # failures(x) px q(n-x) P(x)=n(x)*px*q(n-x) ȹ x Ⱥ 0 1 2 1 4000 7998000 1 0.018279 0.001 0.018297 0.000001 0.018316 0.018279 0.073189 0.146488 P(0 ­2) = 0.0183 + 0.0732 + 0.1465 =0.2380 P(3 or more) = 1 – P(0 ­2) = 1 ­0.2380 = 0.7620 b) The expected value of the binomial distribution is n p = 4000 x 0.001 = 4...
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