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An aircraft seam requires 22 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 19% 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 9% of all seams need reworking?

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zainsubhani

Answer:

Part A:

[tex]p=0.0095[/tex]

Part B:

[tex]p=0.0043[/tex]

Step-by-step explanation:

Part A:

The number of rivets=22 rivets

Probability that no rivet is defective= (1-p)^22

The probability that at least one rivet is defective=1-(1-p)^22

For 19% of all seams need reworking, probability that a rivet is defective is given by:

[tex]1-(1-p)^{22}=0.19[/tex]

[tex](1-p)^{22}=0.81\\p=1-\sqrt[22]{0.81} \\p=0.0095[/tex]

Part B:

For 9% of all seams need reworking, probability of a defective rivet is:

[tex]1-(1-p)^{22}=0.09\\p=1-\sqrt[22]{0.91} \\p=0.0043[/tex]

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