A committee of two persons is selected from two men and two women.

Find the probability that the committee will have at least one man.

total number of choices

= C(4,2) = 6

number of ways to pick 2 women
= C(2,2) = 1

number of ways with at least 1 man
= 1 - 1/6
= 5/6

Proof:
let the women be A, and B
let the men be C and D

choices
AB AC AD
BC BD
CD
of those 5 cases contain A or B
only one does not contain either an A or a B
My answer is correct

To find the probability that the committee will have at least one man, we need to calculate the total number of possible committees as well as the number of committees that have at least one man.

First, let's determine the total number of possible committees. Since we are selecting a committee of two persons from two men and two women, we can use the combination formula.

The total number of possible committees (without any restrictions) can be calculated as:
Total number of possible committees = C(2 men, 2 committee spots) * C(2 women, 0 committee spots)
+ C(2 men, 1 committee spot) * C(2 women, 1 committee spot)
+ C(2 men, 0 committee spots) * C(2 women, 2 committee spots)

To calculate the number of committees that have at least one man, we need to subtract the number of committees with no men from the total number of committees.

The number of committees with no men can be calculated as:
Number of committees with no men = C(2 women, 2 committee spots)

Finally, we can use the formula:
Probability = (Total number of committees - Number of committees with no men) / Total number of possible committees

Let's calculate the solution step by step:

1. Total number of possible committees = C(2, 2) * C(2, 0) + C(2, 1) * C(2, 1) + C(2, 0) * C(2, 2) = 1 * 1 + 2 * 2 + 1 * 1 = 1 + 4 + 1 = 6

2. Number of committees with no men = C(2, 2) = 1

3. Probability = (Total number of committees - Number of committees with no men) / Total number of possible committees = (6 - 1) / 6 = 5 / 6

Therefore, the probability that the committee will have at least one man is 5/6.