Question

An aircraft seam requires 30 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 21% 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 11% of all seams need reworking?

Answer 1

a. 0.00783

b. 0.003876

Explanation:

The computation is shown below;

a. The probability for the rivet to be defective is

Let us assume A is the event for seam failure and B would be event for rivets failure

Now

a)
`P[A] = 1 - P[B']^(30)`

`0.21 = 1 - P[B']^(30)`

`0.79 = P[B']^(30)`

`P[B'] = 0.79^{(1)/(30)}`

P[B'] = 0.99217

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

= 0.00783

b) Now the Next one is

`0.08 = 1 - P[B']^(25)`

`0.89 =P[B']^(30)`

`P[B'] = 0.89^{((1)/(30))}`

= 0.99612

So,

P[B] is

= 1 - P[B']

= 0.003876

We simply applied the above formula so that each one part could be calculated i.e the probabilities of the given question