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mgmat ds-3

by pradeepkaushal9518 » Sat Sep 11, 2010 2:10 am
If a jury of 12 people is to be selected randomly from a pool of 15 potential jurors, and the jury pool consists of 2/3 men and 1/3 women, what is the probability that the jury will comprise at least 2/3 men?
24/91
45/91
2/3
67/91
84/91
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by Rahul@gurome » Sat Sep 11, 2010 3:32 am
2/3 are men implies number of men = (2/3)*15 = 10 and number of women = 5
We have to find the probability that the jury will comprise at least 2/3 men or (2/3) of 12 = 8 men
So, the possible combinations are: (8M, 4W), (9M, 3W), (10M, 2W), where M is the number of men and W is the number of women.

Required probability = [(10C8 * 5C4) + (10C9 * 5C3) + (10C10 * 5C2)]/[15C12]
= [(45*5) + (10*10) + (1*10)]/[35*13]
= [225 + 100 + 10]/455
= 335/455 = 67/91

The correct answer is [spoiler](D)[/spoiler].
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