In a group of 7 men and 8 women, how many committees of 5 can be selected for which there are 2 men and 3 women? What is the probability of selecting a committee of 5 consisting of 2 men and 3 women?
C(7,2)*C(8,3) ways of choosing 2 out of 7 men and 3 out of 8 women. C(n,k) = n!/[(k!)(n-k)!]
C(15,5) ways of picking 5 committee members out of 15
probability of selecting 2 men and 3 women is good over total --
C(7,2)*C(8,3)/C(15,5)
To find out how many committees of 5 can be selected with 2 men and 3 women from a group of 7 men and 8 women, we can use combinations.
The number of ways to select 2 men from a group of 7 men is given by the combination formula: C(7, 2) = 7! / (2! * (7 - 2)!) = 21.
Similarly, the number of ways to select 3 women from a group of 8 women is: C(8, 3) = 8! / (3! * (8 - 3)!) = 56.
Since the selection of men and women are independent selections, we can multiply the number of ways to select men and women together to get the total number of committees: 21 * 56 = 1,176 committees.
Now, to find the probability of selecting a committee of 5 with 2 men and 3 women, we need to divide the number of favorable outcomes (committees with 2 men and 3 women) by the total number of possible outcomes (total number of committees).
The number of committees with 2 men and 3 women is 1,176 (as we calculated earlier).
The total number of possible committees of 5 from a group of 7 men and 8 women is given by the combination formula as well: C(15, 5) = 15! / (5! * (15 - 5)!) = 3003.
Therefore, the probability of selecting a committee of 5 consisting of 2 men and 3 women is: 1,176 / 3003 ≈ 0.391 or approximately 39.1%.