Question

Ninety percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive vehicles pass or fail independently of one another, what is the probability that at least one of the next six inspected fails?

Answer 1

46.86% probability that at least one of the next six inspected fails

Explanation:

For each vehicle, there are only two possible outcomes. Either it passes the inspection, or it does not. The probability of a vehicle passing the inspection is independent of other vehicles. 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.

Ninety percent of all vehicles examined at a certain emissions inspection station pass the inspection.

100 - 90 = 10% fail, so
`p = 0.1`

Six vehicles:

This means that
`n = 6`

What is the probability that at least one of the next six inspected fails?

Either none fail, or at least one of them does. The sum of the probabilities of these outcomes is 1. So

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

We want
`P(X \geq 1)`

Then

`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_(6,0).(0.1)^(0).(0.9)^(6) = 0.5314`

`P(X \geq 1) = 1 - P(X = 0) = 1 - 0.5314 = 0.4686`

46.86% probability that at least one of the next six inspected fails