Question

A manager of a store determines that there is a 0.22 probability that a randomly selected customer who enters the store will make a purchase. If 20 customers enter the store and their decision to make a purchase is independent of the other customers’ decisions, what is the probability that at least one of the customers makes a purchase? Round your answer to 3 decimal places.

Answer 1

Answer:

0.993 = 99.3% probability that at least one of the customers makes a purchase

Step-by-step explanation:

For each customer, there are only two possible outcomes. Either they make a purchase, or they do not. The probability of a customer making a purchase is independent of any 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.

0.22 probability that a randomly selected customer who enters the store will make a purchase.

This means that $p = 0.22$

20 customers

This means that $n = 20$

What is the probability that at least one of the customers makes a purchase?

This is

$P(X \geq 1) = 1 - P(X = 0)$

In which

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

$P(X = 0) = C_{20,0}.(0.22)^{0}.(0.78)^{20} = 0.007$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.007 = 0.993$

0.993 = 99.3% probability that at least one of the customers makes a purchase

Answer 2

Answer:

.993

Step-by-step explanation: