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delta09
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 woman
So 1 women or no women..
No women = 11C4..
1 women (from A) = 7C1*11C3..
ans = \(11C4+7C1*11C3\)
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