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Jury members selection- MGMAT

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Manager
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Jury members selection- MGMAT [#permalink]  20 May 2009, 15:37
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Question Stats:

40% (02:56) correct 60% (09:54) 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 will comprise at least 2/3 men?

A 24/91
B 45/91
C 2/3
D 67/91
E 84/91

Can anyone give a detailed explanation for this problem?
Thanks
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Re: Jury members selection- MGMAT [#permalink]  09 Aug 2010, 13:02
1
KUDOS
I would apppreciate kudos

It is very easy . I have solved this in less than a minute.

there are total 10 men and 5 women
in 12 jury members, could be:

M W T
10+2=12
9+3=12
8+4=12
7+5=12
M-men , W-women T - total

So out of 4 possible outcomes 3 are favorbale, 1 is unfavorable, it is 3/4=0.75

67/91~73%., which is the closest to 3/4.

Bingo!
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Director
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Re: Jury members selection- MGMAT [#permalink]  21 May 2009, 02:27
Hi,

You have 10 men and 5 women in the initial jury. The question asks that you should have at least 2/3 men which means that in 12 people selected at least 8 are men.

But if you select 12 out of 15 you can do it in (15!)/(12!*3!) or 455 ways. Note that the worst option i.e. with least men is 7men 5 women which can be done in 120 ways.

All other options are OK meaning 8m 4W, 9M 3W etc.

Solution is 1- 120/455=335/455=67/91

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Re: Jury members selection- MGMAT [#permalink]  21 May 2009, 18:15
BG wrote:
with least men is 7men 5 women which can be done in 120 ways.

Can you please explain how you got the 120 ways?
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Re: Jury members selection- MGMAT [#permalink]  21 May 2009, 23:34
Hi,

selecting 5 women can be done in one way because all the women are 5. Now selecting 7 men out of 10 can be done in 120 ways ( 10!/(7!*3!))

So there are 120 options with 7 men and 5 women

regards
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Re: Jury members selection- MGMAT [#permalink]  22 May 2009, 04:17
Mine approach:

Find out the probablity to select 3 men ( 1/3 of men)

10C3/15C12=24/91

Hence probability that the jury will comprise at least 2/3 men = 1-24/91=67/91
Hence D.
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Re: Jury members selection- MGMAT [#permalink]  10 Jun 2009, 10:41
i should be able to also solve this by summing the different combinations right ? I.e. 8m2w + 9m1w + 10m .... so (10C8)(5C2) + (10C9)(5C1) + (10C10) ... and then dividing by 15C12 ?
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Re: Jury members selection- MGMAT [#permalink]  10 Jun 2009, 15:41
There can be below scenarios.
a). 8M + 4W = 10C8 5C4 = 225 ways.
b). 9M + 3W = 10C9 5C3 = 100
c). 10M + 2W = 10C10 5C2 = 10
So total selections possible = a+b+c = 335
Total ways to select 12 = 15C12 = 455

Probability = 335/455 = 67/91.
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Re: Jury members selection- MGMAT [#permalink]  27 Feb 2011, 19:35
This question took a lot of time on a recent test that i took! Is your method sound? It is very crude but it does give you the answer. I was wondering if i can extend it to other similar qns.
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Re: Jury members selection- MGMAT [#permalink]  27 Feb 2011, 23:40
mbafall2011 wrote:
This question took a lot of time on a recent test that i took! Is your method sound? It is very crude but it does give you the answer. I was wondering if i can extend it to other similar qns.

i think yes.
As long as the the way you solve the question is logical you may apply any technique that best fits you.
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Manager
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Re: Jury members selection- MGMAT [#permalink]  28 Feb 2011, 03:34
There are 10 Men & 5 Females

Possible choices :
10M & 2F
9M & 3F
8M & 4F
7M & 5F

so :
P(at least 2/3 males = 8 males at least)
= 1 - P(7males and 5 females)
= 1 - [C(10,7)*C(5,5)/C(15,12)]
= 1 - 120*1/455
= 335/455
= 67/91

OA : D
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Re: Jury members selection- MGMAT [#permalink]  14 Sep 2011, 09:46
12 people to be selected from a pool of 10 men and 5 women. Find the probability that the jury will have at least 8 men (means 8 men, 4 women or 9 men, 3 women... etc.). In other words, we can find the probability of having 5 women out of the pool of 5 women and 7 men out of the pool of 10. Then subtract this by 1. You should get 67/91
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Re: Jury members selection- MGMAT [#permalink]  19 Dec 2011, 06:32
My approach:

P=\frac{C^10_8*C^5_4+C^10_9*C^5_3+C^10_10*C^5_2}{C^15_12}=\frac{67}{91}
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Re: Jury members selection- MGMAT [#permalink]  20 Dec 2011, 06:22
nice explained by mdfrahim
my vote for 67/91
Re: Jury members selection- MGMAT   [#permalink] 20 Dec 2011, 06:22
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