bhumika.k.shah wrote:From a group of 6 men and 3 women a committee of four individuals is
selected. What is the probability that the committee:

a) Includes 2 women and 2 men?
b) Includes at most one woman?
c) Includes at least one woman?
d) Includes at least one man?

Again a detailed explanation

a) From a group of 6 men 2 men can be selected in 6C2 ways; From a group of 3 women 2 women can be selected in 3C2 ways

Total no of ways = 6C2*3C2 = 15*3 =45

b) Committe can be formed with 4 men or 3 men and 1 women

4 Men in 6C4 ways =15
3 men and 1 women in 6C3*3C1 =20*3 =60

total 75 ways

c) 1 women 3 men, 2 women 2 men, or 3 women 1 man

60+45+6C1*3C3 = 111 ways

d) 1 Man 3 women, 2 women 2 men, 3 Men 1 woman, 4 Men

6+45+60+15

126 ways

bhumika.k.shah wrote:From a group of 6 men and 3 women a committee of four individuals is
selected. What is the probability that the committee:

a) Includes 2 women and 2 men?
b) Includes at most one woman?
c) Includes at least one woman?
d) Includes at least one man?

Again a detailed explanation

Just to finish the question:

probability = # of desired outcomes/total # of possibilities

We're selecting 4 out of 9 people, so total # of possibilities is:

9C4 = 9!/4!5! = 9*8*7*6/4*3*2 = 9*8*7/4 = 9*2*7 = 126

So, using ajith's beautiful calculations, the answers are:

a) 45/126 = 5/14
b) 75/126 = 25/42
c) 111/126 = 37/42
d) 100%

We didn't actually have to do any calculations for d; there are only 3 women available, so every possible selection of 4 people will include at least 1 man.

Thanks u guys!
i finally got this sum without any doubts