.A committee of 6 is to be selected from a group of 6 men and 9 women. If the selection is made randomly, what is the probability that the committee consists of 2 men and 3 women?
number of selections without conditions
= C(15,6)
number of selections as stated
= C(6,2) x C(9,3)
Problem with your question .....
2 men and 3 women adds up to only 5 people chosen, but you are choosing 6.
Fix your problem and I will continue
To find the probability of selecting 2 men and 3 women out of a group of 6 men and 9 women, we need to determine the total number of ways of selecting a committee and the number of favorable outcomes.
Total number of ways to select a committee:
The total number of ways to select a committee of 6 members from a group of 15 people (6 men + 9 women) is given by the combination formula. We use the combination formula because the order in which the committee members are selected does not matter.
The formula for the combination is nCr, where n is the total number of people and r is the number of committee members to be selected.
In this case, we have n = 15 and r = 6, so the total number of ways to select a committee is:
15C6 = 15! / (6! * (15-6)!) = (15 * 14 * 13 * 12 * 11 * 10) / (6 * 5 * 4 * 3 * 2 * 1) = 5005
Number of favorable outcomes:
To determine the number of favorable outcomes, we need to find the number of ways to select 2 men from 6 men and 3 women from 9 women.
The number of ways to select 2 men from 6 men is given by the combination formula:
6C2 = 6! / (2! * (6-2)!) = (6 * 5) / (2 * 1) = 15
Similarly, the number of ways to select 3 women from 9 women is:
9C3 = 9! / (3! * (9-3)!) = (9 * 8 * 7) / (3 * 2 * 1) = 84
So, the number of favorable outcomes is 15 * 84 = 1260.
Probability:
Finally, we can find the probability of selecting 2 men and 3 women by dividing the number of favorable outcomes by the total number of ways to select a committee:
Probability = Number of favorable outcomes / Total number of ways to select a committee
Probability = 1260 / 5005 ≈ 0.251