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If a jury of 12 people is to be selected randomly from a [#permalink]

17 Aug 2010, 17:41

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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. 5/91
c. 2/3
d. 67/91
e. 84/91

I'm lost on the best approach to solve this question. Any help would be appreciated.
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Re: Jury Duty [#permalink]

17 Aug 2010, 17:48

If the jury were to be 7M and 5W then the combination is 10C7 5C5. Anyother combination will necessarily have >7M and <5W which is what the question is asking as 2/3 of 12 = 8

Hence the prob = 1 - 10C7/15C12 =24/91
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Manager

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Re: Jury Duty [#permalink]

17 Aug 2010, 17:56

I'm sorry, but what do the variables mean?
10C7 5C5

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Re: Jury Duty [#permalink]

17 Aug 2010, 19:18

estreet wrote:

Answer should be 67/91.

At least 2/3 men would be 8 men, 4 women or 9 men, 3 women or 10 men, 2 women.

Hence total outcomes or combination or picking up these is 15C12.

Denominator would be (13 * 7 * 5)

8 men, 4 women combination is 10C8 * 5C4 -- (45 * 5)

9 men, 3 women combination is 10C9 * 5C3 -- (10 * 10)

10 men, 2 women combination is 10C10 * 5C2 -- (10)

Hence total probability -- ((45 * 5) + (10 * 10) + (10))/(13 * 7 *5)

=> (45 + 20 + 2) / (13 * 7)

=> (67) / (91) -- Answer choice D.
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Senior Manager

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Re: Jury Duty [#permalink]

17 Aug 2010, 19:21

estreet wrote:

I'm sorry, but what do the variables mean?
10C7 5C5

10C7 -- total ways of picking 7 items out of 10 available items (combination) and 5C5 -- ways of picking 5 items out of 5 available items.
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Re: Jury Duty [#permalink]

09 Oct 2010, 15:22

I also used ezhilkumarank's method. But, using a 1-p(7 or less men) may be less time consuming
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Re: Jury Duty [#permalink] 09 Oct 2010, 15:22

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