Question

An aircraft seam requires 24 rivets. The seam will have to be reworked if any of these rivets are defective. Suppose rivets are defective independently of one another, each with the same probability. (Round your answers to four decimal places.)
a. If 16% 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 8% of all seams need reworking?

Answer 1

Answer:

Step-by-step explanation:

a)

p(d) - probability of a rivet being defective

p(r) - probability of seam needing to be reworked

The probability of the seem needing to be reworked is equal to the probability that ANY of 24 rivets is defective

Thus the probability that none of the 25 rivets is defective is 1-p(r)

p(r) = 16%, thus 1-p(r) = 84%

1-p(r) = (1-p(d))^24

0.84 = (1-p(d))^24

0.84^(1/24) = 1-p(d)

==> 1-p(d) = 0.9927615998

==> p(d) = 1-0.9927

==> p(d) = 0.0073

b) given p(r) = 8%, thus 1-p(r) = 92%

1-p(r) = (1-p(d))^24

0.92 = (1-p(d))^24

0.92^(1/24) = 1-p(d)

==> 1-p(d) = 0.99653179446

==> p(d) = 1-0.9965

==> p(d) = 0.0035