Question

Twenty-five percent of the customers entering a grocery store between 5 P.M. and 7 P.M. use an express checkout. Consider five randomly selected customers, and let x denote the number among the five who use the express checkout. Find the probability that 2 of the customers use the express checkout.

Answer 1

The required probability = 0.2637

Explanation:

Let X represents the no of customers that uses the express checkout.

`X \sim Binomial (n =5, p= 0.25)`

The probability mass function is given by:

`P(X =x) = (5!)/(x!(5-x)!)(0.25)^x (1-0.25)^(5-x) \ \ ; for \ x =0,1,2,3,4,5`

To find:

`P(X =2) = (5!)/(2!(5-2)!)(0.25)^2 (1-0.25)^(5-2) \ \`

`P(X =2) = (5!)/(2!(3)!)(0.25)^2 (0.75)^(3) \ \`

`P(X =2) = (5* 4 * 3!)/(2!(3)!)(0.25)^2 (0.75)^(3)`

`P(X =2) = (20)/(2!)(0.25)^2 (0.75)^(3)`

P(X = 2) = 0.2637

Thus, the required probability = 0.2637