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# PS: Probability

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PS: Probability [#permalink]  16 Nov 2008, 05:53
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Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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 comprise at least 2/3 men?

a) 24/91

b) 45/91

c) 2/3

d) 67/91

e) 84/91
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Re: PS: Probability [#permalink]  16 Nov 2008, 07:47
Expert's post
D

P=\frac{C^{10}_{8}*C^{5}_{4}+C^{10}_{9}*C^{5}_{3}+C^{10}_{10}*C^{5}_{2}}{C^{15}_{12}}=\frac{\frac{10*9}{2}*5+10*10+1*10}{\frac{15*14*13}{3*2}}=\frac{5*(45+20+2)}{5*91}=\frac{67}{91}
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Re: PS: Probability [#permalink]  16 Nov 2008, 08:08
You have 10 men and 5 women.

Required probability is at least 8 men out of the 12 jurors.

This means we may have 8, 9 or 10 men in the 12 member jury.

Find the probability for each and add them up.

Probability of choosing 8 men and 4 women is:

(10C8*5C4)/15C12

Similarly for 9 and 10: (10C9*5C3)/15C12 and (10C10*5C2)/15C12.

Add them up and you get 67/91.
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Re: PS: Probability [#permalink]  16 Nov 2008, 08:14
IMO = D

Out of 15 juries, there are 10 men and 5 women.

1) 8 men, 4 women = 225 ways
2) 9 men, 3 women = 100 ways
3) 10 men, 2 women = 10 ways

The total number of ways to select is 335 ways out of 455 (15C12 - the number of possible ways) or 67 / 91
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Re: PS: Probability [#permalink]  16 Nov 2008, 10:12
agree with D

edited didnt see 'at least'
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Re: PS: Probability [#permalink]  16 Nov 2008, 12:26
OA is D. Thanks everyone!
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Re: PS: Probability [#permalink]  16 Nov 2008, 13:20
Agree with D Jury and Jury pool are two different things. Initially was reading them the same and then realized why my answer was not matching up
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Re: PS: Probability [#permalink]  16 Nov 2008, 19:59
D
(10c8*5c4 + 10c9*5c3+10c10*5c2)/15c12
Re: PS: Probability   [#permalink] 16 Nov 2008, 19:59
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