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An aircraft seam requires 29 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 23% of all seams need reworking, what is the probability that a rivet is defective

User Qbyte
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Answer:

0.9995 = 99.95% probability that a rivet is defective

Explanation:

For each rivet, there is only two possible outcomes. Either they are defective, or they are not. Rivets are defective independently of one another, each with the same probability, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.


P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)

In which
C_(n,x) is the number of different combinations of x objects from a set of n elements, given by the following formula.


C_(n,x) = (n!)/(x!(n-x)!)

And p is the probability of X happening.

An aircraft seam requires 29 rivets.

This means that
n = 29

23% of all seams need reworking

This means that
p = 0.23

What is the probability that a rivet is defective?

This is:


P(X \geq 1) = 1 - P(X = 0)

In which


P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)


P(X = 0) = C_(29,0).(0.23)^(0).(0.77)^(29) = 0.0005

Then


P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0005 = 0.9995

0.9995 = 99.95% probability that a rivet is defective

User Karen Tsirunyan
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