Answer to Question #215634 in English for Abid Ali

Question

A class contains 10 men 20 women of which half of men and half of the women have brown eyes a person is selected one by one find the probability that

First one has brown eye and second and third are not brown eyes

First two are women and third and forth are men

Answer 1

We are given that the class contains 10 men and 20 women of which half of men(5 men) and half of the women(10 women) have brown eyes and the total number of people in the class is 30. The number of men with brown eyes is 5 The number of women with brown eyes is 10.

Let P(A) be the event that it is a man (10 out of 30)

Let P(B) be that the person has brown eyes (5 men and 10 woman - 15 out of 30)

( P(A∩B) is a man AND has brown eyes 5/30)

P(A∪B) = P(A) + P(B) - P(A∩B)

= 10/30 + 15/30 - 5/30

=2/3

that is

(10/30*5/30)+(10/30*5/30)+(20/30*10/30)

=0.882

therefore the probability is as done above.