Question

An offline retail store called CoolStore have customers that are either students or not. Customers can buy a plastic bag or not. We know for a fact that: Twenty-five percent of all customers are students. Among customers that are students, eighty percent buy a plastic bag. Among customers who buy a plastic bag, fifty percent are students. Forty customers are chosen at random. The probability is one percent that the number of customers buying a plastic bag exceeds what number?

Answer 1

let A be the event that a customer is student then A' be the event that a customer is not a student.

let X be the event that a customer buys plastic bag then X' be the event that a customer does not buy plastic.

as stated in the problem,

P(A)=0.25

P(A')=0.75

P(X | A) =0.8

P(A | X) =0.5

then

P(A' | X) = 1 - P(A | X) = 1 - 0.5 =0.5

and P(X | A)= P(A | X) *P(X) / P(A) [ BY BAYES' THEOREM]

which gives P(X) = P(X | A)*P(A) / P(A | X) = 0.8*0.25 / 0.5 = 0.4

Probability that a random customer buys a plastic bag = P(X) = 0.4

now, number of customers = n = 40

success probability = p = P(X) = 0.4

This is a binomial probability case.

We have to find c such that P(X > c) = 0.01

using normal approximation, mean = np = 40*0.4 =16 and standard deviation = sqrt(np(1-p)) = sqrt(40*0.4*0.6) = sqrt(9.6) = 3.0984

so, We have to find c such that P(X > c) = 0.01 which gives P(X < c) = 0.99

for a 99% less than probabilty we need to have a z score of 2.326

so, raw score = X = c = mean + z * standard deviation = 16 + 2.326*3.0984 = 23.207

The probability is one percent that the number of customers buying a plastic bag exceeds 23.207 customers out of 40 selected customers.