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 20% 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?

b) As the probability of a seam not to be reworked (good seam) means that all 25 rivets are not defective, as each probability is independent from others, we multiply the probability of a rivet not being defective (RND) 25 times:

[tex]P_{\mbox{Good seam}}=( P_{\mbox{RND}} )^{25}[/tex]

We look up what is the probability of a rivet not being defective because it's easier to do that and knowing that ([tex]P_{D}[/tex] is the probability of a rivet being defective):

[tex]P_{\mbox{RND}} +P_{\mbox{D}}=1[/tex]

We know that the probability of a seam being good is 80% (the same principle as above, 100%-20%=80%), so we can find the probability of a rivet not being defective:

[tex]\sqrt[25]{P_{\mbox{Good seam}}} =P_{\mbox{RND}}\\\\\sqrt[25]{0.8}= P_{\mbox{RND}}=0.99111[/tex]

We can find now the probability of a defective rivet:

[tex]P_{D}=1- P_{\mbox{RND}}\\P_{D}=1-0.99111=8,886*10^{-3}[/tex]

b) Now, the probability of a good seam is 0.9 (100%-10%=90%), we proceed the same as the previous point:

[tex]\sqrt[25]{P_{\mbox{Good seam}}} =P_{\mbox{RND}}\\\\\sqrt[25]{0.9}= P_{\mbox{RND}}=0.99579[/tex]

[tex]P_{D}=1- P_{\mbox{RND}}\\P_{D}=1-0.99579=4.206*10^{-3}[/tex]