Question

Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 60% can be repaired, whereas the other 40% must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly four will end up being replaced under warranty

Answer 1

Answer:

0.0052 = 0.52% probability that exactly four will end up being replaced under warranty.

Step-by-step explanation:

For each telephone, there are only two possible outcomes. Either they are replaced under warranty, or they are not. The probability of a telephone being replaced under warranty is independent of any other telephone, 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.

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

And p is the probability of X happening.

Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of those, 40% are replaced.

This means that $p = 0.2*0.4 = 0.08$

Company purchases ten of these telephones.

This means that $n = 10$

What is the probability that exactly four will end up being replaced under warranty?

This is P(X = 4). So

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

$P(X = 4) = C_{10,4}.(0.08)^{4}.(0.92)^{6} = 0.0052$

0.0052 = 0.52% probability that exactly four will end up being replaced under warranty.