Question

it is found that 4% of watches produced at a particular factory are defective. If 20 watches made at this factory are randomly selected, what is the probbility that at mpst 1 watch in the same sample is found to be defective

Answer 1

0.81 = 81% probability that at most 1 watch in the sample is defective.

Explanation:

For each watch, there are only two possible outcomes. Either it is defective, or it is not. The probability of a watch being defective is independent of any other watch, 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.

4% of watches produced at a particular factory are defective.

This means that
`p = 0.04`

20 watches made at this factory are randomly selected

This means that
`n = 20`

What is the probability that at most 1 watch in the same sample is found to be defective?

This is:

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

In which

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

`P(X = 0) = C_(20,0).(0.04)^(0).(0.96)^(20) = 0.442`

`P(X = 1) = C_(20,1).(0.04)^(1).(0.96)^(19) = 0.368`

Then

`P(X \leq 1) = P(X = 0) + P(X = 1) = 0.442 + 0.368 = 0.81`

0.81 = 81% probability that at most 1 watch in the sample is defective.