An aircraft seam requires 25 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. a. If 15% of all seams need reworking, what is the probability that a rivet is defective? b. How small should the probability of a defective rivet be to ensure that only 10% of all seams need reworking?

Answer: Step 1: Given, An aircraft seam requires 25 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. Number of rivets n = 25. The seam will have to be reworked if any of these rivets is defective and the rivets are defective independently of one another. Step 2: a). To find the probability that a rivet is defective. 15% of all seams need reworking. Probability of seem need to reworked is 100 = 0.15. Let ‘p’ be the probability of rivet defective. Then probability of a not defective is q = 1-p. Therefore, P( seem have to be reworked) = 1- P( seem have not to be reworked) = 1- [(1-p) (1-p).....15 times] = 1 (1 p) 15 Given that probability of seem to be reworked is 0.15. So, 15 1 (1 p) = 0.15 1 - 0.15 = (1 p) 15 15 0.85 = (1 p) 15 Then, (1-p) = (0.85) = 0.9892 P = 1 - 0.9892 P = 0.0108. Therefore, Probability that a rivet defective is 0.0108.