Question

The probability that a certain make of car will need repairs in the first seven months is 0.9. A dealer sells three such cars. What is the probability that at least one of them will require repairs in the first seven months? Round your final answer to four decimal places.

Answer 1

Answer:

0.9990 = 99.90% probability that at least one of them will require repairs in the first seven months.

Step-by-step explanation:

For each car, there are only two possible outcomes. Either they will require repair in the first seven months, or they will not. The probability of a car requiring repair in the first seven months 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The probability that a certain make of car will need repairs in the first seven months is 0.9.

This means that $p = 0.9$A dealer sells three such cars.

A dealer sells three such cars.

This means that $n = 3$

What is the probability that at least one of them will require repairs in the first seven months?

Either none will require repairs, or at least one will. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X > 0) = 1$

We want P(X > 0). So

$P(X > 0) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.9)^{0}.(0.1)^{3} = 0.001$

Then

$P(X > 0) = 1 - P(X = 0) = 1 - 0.001 = 0.9990$

0.9990 = 99.90% probability that at least one of them will require repairs in the first seven months.