Question

A manufacturer of window frames knows from long experience that 5% of the production will have some type of minor defect that will require adjustment. What is the probability that in a sample of 20 window frames, at least one will need adjustment?

Answer 1

64.15% probability that in a sample of 20 window frames, at least one will need adjustment

Explanation:

For each window frame, there are only two possible outcomes. Either it needs adjustment, or it does not. The probability of a window frame needing adjustment is independent of other window frames. So we use the binomial probability distribution 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.

5% of the production will have some type of minor defect that will require adjustment.

This means that
`p = 0.05`

What is the probability that in a sample of 20 window frames, at least one will need adjustment?

This is
`P(X \geq 1)` when n = 20.

We know the either no windows need adjustment, or at least one does. The sum of the probabilities of these events is decimal 1. So

`P(X = 0) + P(X \geq 1) = 1`

`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_(20,0).(0.05)^(0).(0.95)^(20) = 0.3585`

`P(X \geq 1) = 1 - P(X = 0) = 1 - 0.3585 = 0.6415`

64.15% probability that in a sample of 20 window frames, at least one will need adjustment