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SVP
Joined: 21 Jul 2006
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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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CEO
Joined: 17 Nov 2007
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Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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Re: PS: Probability [#permalink]
 16 Nov 2008, 08:47
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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Intern
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Re: PS: Probability [#permalink]
16 Nov 2008, 09:08
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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Intern
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Re: PS: Probability [#permalink]
16 Nov 2008, 09:14
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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Senior Manager
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Re: PS: Probability [#permalink]
16 Nov 2008, 11:12
agree with D edited didnt see 'at least'
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SVP
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Re: PS: Probability [#permalink]
16 Nov 2008, 13:26
OA is D. Thanks everyone!
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VP
Joined: 05 Jul 2008
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Re: PS: Probability [#permalink]
16 Nov 2008, 14:20
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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Senior Manager
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Re: PS: Probability [#permalink]
16 Nov 2008, 20:59
D
(10c8*5c4 + 10c9*5c3+10c10*5c2)/15c12
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Re: PS: Probability
[#permalink]
16 Nov 2008, 20:59
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