An aircraft seam requires 27 rivets The seam will have to be

An aircraft seam requires 27 rivets. The seam will have to be reworked if any of these rivets is defective. Suppose rivets are defective independently of one another, each wi the same probability. (Round your answers to four decimal places.) If 15% of all seams need reworking, what is the probility that a rivet is defective. How small should the probability of a defective rivet be to ensure that only 7% of all seams need reworking?

Solution

a)

Let p = the probability that a rivet is defective

Then

P(at least one defective) = 1 - P(no defective) = 0.15

Hence,

1 - (1-p)^27 = 0.15

(1-p)^27 = 0.85

1-p = 0.99399886

p = 0.00600114 [ANSWER]

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b)

Let p = the probability that a rivet is defective

Then

P(at least one defective) = 1 - P(no defective) = 0.07

Hence,

1 - (1-p)^27 = 0.07

(1-p)^27 = 0.93

1-p = 0.997315805

p = 0.002684195 [ANSWER]