An aircraft seam requires 21 rivets. The seam will have to be reworked if any of

Question

An aircraft seam requires 21 rivets. The seam will have to be reworked if any of these rivets is defective. Suppose rivets are defective independently of one another, each with the same probability, (Round your answers to four decimal places.) (a) If 20% of all seams need reworking, what is the probability that a rivet is defective? Hint: This question essentially asks you to reverse the process that was shown in class. You are given that the chance that all 21 rivets are not defective (i.e., the seam is not defective) is 80%, so what is the chance that any one rivet is not defective? Then what is the chance that any one rivet is defectiv (b) How small should the probability of a defective rivet be to ensure that only 5% of all seams need reworking?

Explanation / Answer

(a) Let the probability that a rivet is defective be p

=> Probability that a rivet is not defective = 1 - p

=> Probability that none of the 21 rivets are defective = (1 - p)21

=> Probability that atleast one rivet is defective (i.e the seam needs to be reworked) = 1 - (1 - p)21

Given 1 - (1 - p)21 = 0.2

=> (1 - p)21 = 0.8

=> (1 - p) = 0.81/21 = 0.9894

=> p = 1 - 0.9894 = 0.0106

(b) 1 - (1 - p)21 = 0.05

=> (1 - p)21 = 0.95

=> (1 - p) = 0.951/21 = 0.9976

=> p = 1 - 0.9976 = 0.0024.

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