# Conditional probability given 2 scenarios: probability of meeting male and a female in the shop

#### Collins

##### New member
There are 4 books being sold in the bookshop : A, B, C, D.

We know that 20% of the male customers buy book A at least once a week, 55% buy book B at least once a week, 25% buy book C at least once a week and 15% buy book D at least once in a month.

We also know that 32% of the female customers by book A at least once a week, 80% buy book B at least once a week, 40% buy book C at least once a week and 65% buy book D at least once a week.

The ratio of male customers to female is 3 to 1.

The goal is to calculate a probability of meeting male and a female in the shop, given that each customer decided to purchase books A, B, C and the average frequency of shopping is once a week.

I believe the solution is to calculate joint probability of male and female probabilities of buying ABC set. Maybe I'm wrong so I could use some help. Also I'm not sure if shopping frequency matters.

#### Country Boy

##### Well-known member
MHB Math Helper

Yes, "the average frequency of shopping is once a week" matters because we are told how many of books A, B, and C are bought a week. (We are told the number of books, D, are bought in a month, but the question doesn't ask about "D".)

Imagine 10000 customers. "The ratio of male customers to female is 3 to 1." So there are (3/4)(10000)= 7500 male customers and 2500 female customers.

"We know that 20% of the male customers buy book A at least once a week, 55% buy book B at least once a week, 25% buy book C at least once a week and 15% buy book D at least once in a month."
So of the male customers 0.20(7500)= 1500 buy book A, 0.55(7500)= 4125 buy book B, 0.25(7500)= 1875 buy book C, and 0.15(7500)= 1125 buy book D.

"We also know that 32% of the female customers buy book A at least once a week, 80% buy book B at least once a week, 40% buy book C at least once a week and 65% buy book D at least once a week."
So of the female customers 0.32(2500)= 800 buy book A, 0.80(2500)= 2000 buy book B, .4(2500)= 1000 buy book C, and .65(2500)= 1625 buy book D.

A total of 1500+ 800= 2300 buy book A, 1500 of them men, 800 or them women. Given that a person buys book A the probability the person is male is $$\frac{1500}{2300}= 0.652$$ (rounded) and the probability the person is female is $$\frac{800}{2300}= 0.348$$. Of course, 0.348= 1- 0.652.

A total of 4125+ 2000= 6125 buy book B, 4125 of them men, 2000 of them women . Given that a person buys book B the probability the person is male is $$\frac{4125}{6125}= 0.673 and the probability the person is female is [tex]\frac{2000}{6125}= 0.327$$.

A total of 1875+ 1000= 2875 buy book C, 1875 of them men, 1000 of them women . Given that a person buys book C the probability the person is male is $$\frac{1875}{2875}= 0.652 and the probability the person is female is [tex]\frac{1000}{2875}= 0.348$$.

If you meant that, of the men, "15% buy book D at least once in a week" rather than "month" and intended to include people who bought book D, then a total of 1125+ 1625= 2750 buy book D, 1125 of them men, 1625 of them women. Given that a person buys book D, the probability the customer is a man is $$\frac{1125}{2750}= 0.409$$ and the probability the customer is a woman is $$\frac{1625}{2750}= 0.591$$.