Question

A manufacturing machine has a 10% defect rate. If 3 items are chosen at random, what is the probability that at least one will have a defect

Answer 1

0.271 = 27.1% probability that at least one will have a defect

Explanation:

For each item, there are only two possible outcomes. Either they have a defect, or they do not. Items are independent. This means that 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.

A manufacturing machine has a 10% defect rate.

This means that
`p = 0.1`

3 items are chosen at random

This means that
`n = 3`

What is the probability that at least one will have a defect?

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_(3,0).(0.1)^(0).(0.9)^(3) = 0.729`

`P(X \geq 1) = 1 - P(X = 0) = 1 - 0.729 = 0.271`

0.271 = 27.1% probability that at least one will have a defect