Question

A consumer organization estimates that over a​ 1-year period 17​% of cars will need to be repaired​ once, 10​% will need repairs​ twice, and 2​% will require three or more repairs. If you own two​ cars, what is the probability that ​a) neither will need​ repair? ​b) both will need​ repair? ​c) at least one car will need​ repair?

Answer 1

a) 50.41% probability that neither will need repair.

b) 8.41% probability that both will need repair.

c) 49.59% probability that at least one car will need repair.

Explanation:

For each car, there are only two possible outcomes. Either it will need repairs, or it will not need repairs. The probability of a car needing repairs is independent of other cars. 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.

17​% of cars will need to be repaired​ once, 10​% will need repairs​ twice, and 2​% will require three or more repairs.

This means that
`p = 0.17+0.1+0.02 = 0.29`

Two cars:

This means that
`n = 2`

a) neither will need​ repair?

This is P(X = 0).

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

`P(X = 0) = C_(2,0).(0.29)^(0).(0.71)^(2) = 0.5041`

50.41% probability that neither will need repair.

​b) both will need​ repair? ​

This is P(X = 2).

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

`P(X = 2) = C_(2,2).(0.29)^(2).(0.71)^(0) = 0.0841`

8.41% probability that both will need repair.

c) at least one car will need​ repair?

Either none will need repair, or at least one will. The sum of these probabilities is 100%.

From a)

50.41% probability that neither will need repair.

p = 100 - 50.41 = 49.59%

49.59% probability that at least one car will need repair.