A group of 6 men and 6 women is randomly divided into 2 groups of size 6 each. What is the probability that both groups will have the same number of men?

0. 4329

Step by step explanation

Since both groups will have the same number of men, we would guarantee each group consists of three women and three men.

So theprobability is (6, 3)·(6, 3)/(12, 6)=100/231.

P(A) = 400/924 = 100/231 or 0.4329

Step-by-step explanation:

The probability that both groups will have the same number of men P(A);

For the two groups to have the same number of men they must include 3 men and 3 women in each group.

P(A) = Number of possible selections of 3 men from 6 and 3 women from 6 into each of the two groups N(S) ÷ total number of possible selections of members into the two groups N(T).

P(A) = N(S)/N(T)

Since order is not important, we will use combination.

N(S) = 6C3 × 6C3 = 20 × 20 = 400

N(T) = 12C6 = 924

P(A) = 400/924 = 100/231 or 0.4329

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