Question

Name: An aircraft rivets is defect with the same probability. seam requires 10 rivets. The seam will have to be reworked if any of these ve. Suppose rivets are defective independently of one another, each (a) If 2 5% of all seams need reworking, what is the probability 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 1

Let A be the event a seam fails.
Let B be the event a rivet fails.

P[A] = 1 - P[B']^10
0.25 = 1 - P[B']^10
0.75 = P[B']^10
P[B'] = 0.75^(1/10)
P[B'] = 0.9716
P[B] = 1 - P[B'] = 0.0284

Next one is 0.1 = 1 - P[B']^10

0.9 = P[B']^10

P[B'] = 0.9^(1/10) = 0.9895

P[B] = 1 - P[B'] = 0.0105

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