delta09 wrote:

Find the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains:

A) exactly 1 woman

B) at least 1 woman

C) at most 1 woman

Hi,

the way you should work on each is..

6M, 7W and 5C.... 4 person committe to be made..

A) exactly 1 woman Choose 1 women out of 7 = 7C1..

choose rest 3 out of 6M and 5C = 11C3..

ans \(= 7C1*11C3..\)

B) at least 1 woman Total ways = \((6+7+5)C4 = 18C4\)...

ways NO women is choosen = \((6+5)C4 = 11C4\)...

ways atleast ONE women is choosen = \(18C4 - 11C4\)...

C) at most 1 womanSo 1 women or no women..

No women = 11C4..

1 women (from A) = 7C1*11C3..

ans = \(11C4+7C1*11C3\)

_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372

2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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