The problem asks for the probability that two specific members, A and B, can be on the committee of 5 members chosen from a group of 12 people, where the committee consists of 6 men and 6 women. This suggests that the 12 people are made of 6 men and 6 women.

1. Problem Description

The problem asks for the probability that two specific members, A and B, can be on the committee of 5 members chosen from a group of 12 people, where the committee consists of 6 men and 6 women. This suggests that the 12 people are made of 6 men and 6 women.

2. Solution Steps

The total number of ways to choose a committee of 5 members out of 12 is given by the combination formula:

$C(n, r) = \frac{n!}{r!(n-r)!}$

Where $n$ is the total number of items and $r$ is the number of items to choose.

Total number of ways to choose 5 members from 12:

$C(12, 5) = \frac{12!}{5!7!} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792$

If A and B must be on the committee, we need to choose the remaining 3 members from the remaining 10 people. The number of ways to do this is:

$C(10, 3) = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$

The probability that A and B are both on the committee is the number of ways to choose the remaining 3 members divided by the total number of ways to choose the committee of 5:

$P(A \text{ and } B \text{ are on committee}) = \frac{C(10, 3)}{C(12, 5)} = \frac{120}{792} = \frac{10}{66} = \frac{5}{33}$

3. Final Answer

5/33