From a group of 6 men and 4 women, a committee os 3 is to be selected at random. Find P(at least 2 women)?

Answer 1

There are a total of "10 choose 3" ways to pick the committee:
C(10, 3) = 10 x 9 x 8 / 6 = 240 ways

Now you could pick 1 man and 2 women, or you could pick 3 women.

The ways to pick 1 man and 2 women = "6 choose 1" x "4 choose 2".
C(6, 1) x C(4, 2) = 6 x (4 x 3 / 2) = 6 x 6 = 36 ways

The ways to pick 3 women = "4 choose 3":
C(4, 3) = 4 ways

P(at least 2 women) = [ C(6,1) x C(4,2) + C(4,3) ] / C(10, 3)
P(at least 2 women) = (4 + 36) / 240
P(at least 2 women) = 40 / 240
P(at least 2 women) = 1/6

Answer 2

you have to break this problem down into individual cases:

case #1: 2 women and 1 man

to choose 2 women you have: 4C2
to choose the 1 man you have: 6C1

multiply these together: 4C2 * 6C1

case#2: 3 women

to choose 3 women you simply choose 3 of the 4 women: 4C3

add this answer to the answer you got for case#1:

4C3 + (4C2 * 6C1) = 4 + (6 * 6) = 40

if you were to choose the two women in 4C2 ways, and then pick your remaining person in 8C1 ways, you would have a double count on the women. the correct answer requires you break the problem into cases, which there can only be two of since you have the restraint that your sub-commttee of 3 people must include AT LEAST 2 women (these allows you to have a sub-committee entirely of women)

hope this helps!

Answer 3

Do your own homework.