Jury

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Jury

by j_shreyans » Tue Sep 30, 2014 3:26 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?



A)24/91


B)45/91


C)2/3


D)67/91


E)84/91

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by GMATGuruNY » Tue Sep 30, 2014 3:32 am
j_shreyans wrote: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?

A) 24/91
B) 45/91
C) 2/3
D) 67/91
E) 84/91
Question rephrased: What is the probability that at least 8 men will be selected to serve on the 12-member jury?

P(good outcome) = 1 - P(bad outcome).

Here, a BAD outcome means selecting a jury with FEWER than 8 men.
Of the 10 men and 5 women in the jury pool, 3 people must be selected NOT to serve on the jury.
There is only ONE WAY to select fewer than 8 men FOR the jury:
The 3 people selected NOT to serve on the jury must ALL be men, leaving 7 men and all 5 women to serve on the jury.

P(1st non-juror is a man) = 10/15. (Of the 15 people in the jury pool, 10 are men.)
P(2nd non-juror is a man) = 9/14. (Of the 14 remaining people in the jury pool, 9 are men.)
P(3rd non-juror is a man) = 8/13. (Of the 13 remaining people in the jury pool, 8 are men.)
Since a bad outcome requires that all 3 events happen, we MULTIPLY the fractions:
10/15 * 9/14 * 8/13 = 24/91.

Thus:
P(good outcome) = 1 - 24/91 = 67/91.

The correct answer is D.
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by ProGMAT » Tue Sep 30, 2014 8:58 am
Hi GMATGuru,

Cant we attempt this question by selecting like:
Probability of selecting (8M & 4W) + Prob. (9M & 3W) + Prob. (10M & 2W)!!

Please explain!
Thanks.
GMATGuruNY wrote:
j_shreyans wrote: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?

A) 24/91
B) 45/91
C) 2/3
D) 67/91
E) 84/91
Question rephrased: What is the probability that at least 8 men will be selected to serve on the 12-member jury?

P(good outcome) = 1 - P(bad outcome).

Here, a BAD outcome means selecting a jury with FEWER than 8 men.
Of the 10 men and 5 women in the jury pool, 3 people must be selected NOT to serve on the jury.
There is only ONE WAY to select fewer than 8 men FOR the jury:
The 3 people selected NOT to serve on the jury must ALL be men, leaving 7 men and all 5 women to serve on the jury.

P(1st non-juror is a man) = 10/15. (Of the 15 people in the jury pool, 10 are men.)
P(2nd non-juror is a man) = 9/14. (Of the 14 remaining people in the jury pool, 9 are men.)
P(3rd non-juror is a man) = 8/13. (Of the 13 remaining people in the jury pool, 8 are men.)
Since a bad outcome requires that all 3 events happen, we MULTIPLY the fractions:
10/15 * 9/14 * 8/13 = 24/91.

Thus:
P(good outcome) = 1 - 24/91 = 67/91.

The correct answer is D.

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by [email protected] » Tue Sep 30, 2014 11:32 am
Hi ProGMAT,

Yes, you CAN answer the question by solving the individual calculations that you listed. Here's how....

Since we're selecting 12 people from a group of 15, and the order doesn't matter, we can use the combination formula:

N!/[K!(N-K)!]

15c12 = 15!/[12!(3!)] = 455 possible groups of 12

Now we calculate each of the possible options that fits what we're looking for:

8 men (from 10) and 4 women (from 5) = (10c8)(5c4) = (45)(5) = 225
9 men (from 10) and 3 women (from 5) = (10c9)(5c3) = (10)(10) = 100
10 men (from 10) and 2 women (from 5) = (10c10)(5c2) = (1)(10) = 10

Total options with at least 8 men = 335

Probability = 335/455 = 67/91

Final Answer: D

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