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The transaction history at an electronic goods store indicates that 21 percent of customers purchase the extended warranty when they buy an eligible item. Suppose customers who buy eligible
items are chosen at random one at a time, until one is found who purchased the extended warranty. Let the random variable X represent the number of customers it takes to find one who
purchased the extended warranty. Assume customers' decisions on whether to purchase the extended warranty are independentWhich of the following is closest to the probability that X > 3;
that is, the probability that it takes more than 3 customers who buy an eligible item to find one who purchased the extended warranty?

Respuesta :

Answer:

.131

Step-by-step explanation

(.79)(.79)(.21) = .131

Using the binomial distribution, it is found that there is a 0.493 = 49.3% probability that it takes more than 3 customers who buy an eligible item to find one who purchased the extended warranty.

For each customer, there are only two possible outcomes, either they purchase extended warranty, or they do not. The probability of a customer purchasing extended warranty is independent of any other customer, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 21% of customers purchase extended warranty, hence [tex]p = 0.21[/tex]

The probability that it takes more than 3 customers who buy an eligible item to find one who purchased the extended warranty is the probability that none of the first three buy, which is P(X = 0) when n = 3, hence:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{3,0}.(0.21)^{0}.(0.79)^{3} = 0.493[/tex]

0.493 = 49.3% probability that it takes more than 3 customers who buy an eligible item to find one who purchased the extended warranty.

A similar problem is given at https://brainly.com/question/24863377

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