Thirty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 70% can be repaired, whereas the other 30% must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly three will end up being replaced under warranty? (Round your answer to three decimal places.)

Answer:26.68% probability that exactly three will end up being replaced under warrantyStep-by-step explanation:For each telephone under warranty, there are only two possible outcomes. Either they need to be replaced, or they do not need to be replaced. Each telephone is independent of other telephones. So we use the binomial probability distribution to solve this question.Binomial probability distributionThe binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]And p is the probability of X happening.30% must be replaced with new unitsThis means that [tex]p = 0.3[/tex]If a company purchases ten of these telephones, what is the probability that exactly three will end up being replaced under warrantyThis is [tex]P(X = 3)[/tex] when [tex]n = 10[/tex]. So[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex][tex]P(X = 3) = C_{10,3}.(0.3)^{3}.(0.7)^{7} = 0.2668[/tex]26.68% probability that exactly three will end up being replaced under warranty