Question

A company has 125 personal computers. The probability that any one of them will require repair on a given day is 0.025. To find the probability that exactly 20 of the computers will require repair on a given day, one will use what type of probability distribution

Answer 1

Answer:

Binomial probability

$P(X = 20) = C_{125,20}.(0.025)^{20}.(0.975)^{95}$

Step-by-step explanation:

For each computer, there are only two possible outcomes. Either they fail, or they do not. The probability of a computer failing is independent from the probability of other computers failing. So we use the binomial probability distribution to solve this problem.

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.

In this problem we have that:

$n = 125, p = 0.025$

To find the probability that exactly 20 of the computers will require repair on a given day, one will use what type of probability distribution

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

$P(X = 20) = C_{125,20}.(0.025)^{20}.(0.975)^{95}$