Question

Not all visitors to a certain company's website are customers. In fact, the website administrator estimates that about 5% of all visitors to the website are looking for other websites. Assuming that this estimate is correct, find the probability that, in a random sample of 4 visitors to the website, exactly 2 actually are looking for the website. Round your response to at least three decimal places.

Answer 1

Answer:

0.0135 = 1.35% probability that, in a random sample of 4 visitors to the website, exactly 2 actually are looking for the website.

Step-by-step explanation:

For each visitor of the website, there are only two possible outcomes. Either they are looking for the website, or they are not. The probability of a customer being looking for the website is independent of other customers. This 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.

5% of all visitors to the website are looking for other websites.

So 100 - 5 = 95% are looking for the website, which means that $p = 0.95$

Find the probability that, in a random sample of 4 visitors to the website, exactly 2 actually are looking for the website.

This is P(X = 2) when n = 4. So

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

$P(X = x) = C_{4,2}.(0.95)^{2}.(0.05)^{2} = 0.0135$

0.0135 = 1.35% probability that, in a random sample of 4 visitors to the website, exactly 2 actually are looking for the website.