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here is a tough pb

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by awax22 » Tue Nov 20, 2007 11:46 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?
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by mtr » Tue Nov 20, 2007 12:43 pm
So out of the 15 potential jurors 10 are men and 5 are women. For there to be at least 2/3rds men on the jury of 12, there needs to be 8 or more men. But since there are only 10 men to choose from there can only be 8, 9 and 10. So I think its:

(10C8*5C4+10C9*5C3+10C10*5C2)/15C12 = 67/91 if my math is correct.

Do you have an OA for this one?
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by mtr » Tue Nov 20, 2007 1:06 pm
Actually you can just do what the chances of it not happening are, and thats when all 5 women are picked. So you can do 1-probability of it not happening.

1-((10C7*5C5)/15C12)=67/91
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