Permutation and combination

Question

Five persons are chosen at random from a group of ten persons consisting of four men and six women. Calculate the probabilities that the four persons chosen will:

1. Consist of five women.

2. Consist of two women and three men

Answer 1

This problem uses the fact that, in a finite sample space, and assuming the uniform distribution (there is no reason to assume that any one person is more likely to be chosen than another, i.e. they are chosen "at random"), then

P(A) = n(A)/n(S)

where A is the event of interest, S is the sample space, and n denotes the cardinality of each set (the number of elements it contains).

Our sample space consists of all possible combinations we can form choosing 5 people from 10, and so n(S) = 10C5 = 252

1. If A is the event that we choose 5 women, then n(A) = 6C5 = 6 (of the 6 women, choose 5). Hence P(A) = 6/252
2. If B is the event that the group consists of 2 women and 3 men, then n(B) = (6C2)*(4C3) = 15*4 = 60 (first, choose 2 of the 6 women, AND THEN choose 3 of the 4 men). So P(B) = 60/252