Question 1157466: A committee of three is chosen at random from a group of four women and five men. Find the probability that the committee contains at least one man.
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
P(at least one man) = 1 - P(no men) = 1-P(3 women)
P(3 women and no men) = (C(4,3)*C(5,0))/C(9,3) = (4*1).84 = 4/84 = 1/21
P(at least one man) = 1-1/21 = 20/21
You can get practice calculating probabilities like this by finding the probabilities of choosing 1, 2, or 3 men.
P(1 man) = (C(4,2)*C(5,1))/C(9,3) = (6*5)/84 = 30/84
P(2 men) = (C(4,1)*C(5,2))/C(9,3) = (4*10)/84 = 40/84
P(3 men) = (C(4,0)*C(5,3))/C(9,3) = (1*10)/84 = 10/84
The sum of the probabilities is (4+30+40+10)/84 = 1.
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