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
GMAT online Tutor
close
0% (00:00) correct 
