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

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

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04 Feb 2008, 16:40
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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. 45/91
C. 2/3
D. 67/91
E. 84/91
[Reveal] Spoiler: OA

Last edited by Bunuel on 11 Feb 2014, 00:08, edited 1 time in total.
Edited the question and added the OA.
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04 Feb 2008, 17:32
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We have a pool of 10 men and 5 women
We need a jury of 12 people

What is the probability that the jury will comprise at least 2/3 men?

2/3 of a jury of 12 = 8 men. There will always be at least 8 men on the jury unless all 5 women are selected, in that case there will be 7 men and 5 women.

15C12 = 455 ways to select a jury of 12 from a pool of 15
5C5 = 1 way to select all 5 women
10C7 = 120 ways to select 7 men from a pool of 10
1*120 = 120 ways to select a jury with fewer than 8 men

(455-120)/455 = 335/455 = 67/91

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06 Feb 2008, 15:35
eschn3am wrote:
We have a pool of 10 men and 5 women
We need a jury of 12 people

What is the probability that the jury will comprise at least 2/3 men?

2/3 of a jury of 12 = 8 men. There will always be at least 8 men on the jury unless all 5 women are selected, in that case there will be 7 men and 5 women.

15C12 = 455 ways to select a jury of 12 from a pool of 15
5C5 = 1 way to select all 5 women
10C7 = 120 ways to select 7 men from a pool of 10
1*120 = 120 ways to select a jury with fewer than 8 men

(455-120)/455 = 335/455 = 67/91

clear answer, thanks - one question though, the numerator is 455-120; is that 120 because you have multiplied by the 1? i.e. if there were two ways, would it have been 120*2 leading to 455-240?
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06 Feb 2008, 18:54
gmatraider wrote:
eschn3am wrote:
We have a pool of 10 men and 5 women
We need a jury of 12 people

What is the probability that the jury will comprise at least 2/3 men?

2/3 of a jury of 12 = 8 men. There will always be at least 8 men on the jury unless all 5 women are selected, in that case there will be 7 men and 5 women.

15C12 = 455 ways to select a jury of 12 from a pool of 15
5C5 = 1 way to select all 5 women
10C7 = 120 ways to select 7 men from a pool of 10
1*120 = 120 ways to select a jury with fewer than 8 men

(455-120)/455 = 335/455 = 67/91

clear answer, thanks - one question though, the numerator is 455-120; is that 120 because you have multiplied by the 1? i.e. if there were two ways, would it have been 120*2 leading to 455-240?

The 120 is the number of possibilities where there are FEWER than 2/3 men on the jury. 455 is the number of possible ways to make up this jury. If there are 455 ways total, and 120 of those ways have fewer than 2/3 men...then (455-120) is the number of possible ways where the jury IS at least 2/3 men. We're looking for the probability that the jury will be at least 2/3 men so (455-120)/455 is our equation.
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16 Nov 2008, 08:47
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$$P=\frac{C^{10}_{8}*C^{5}_{4}+C^{10}_{9}*C^{5}_{3}+C^{10}_{10}*C^{5}_{2}}{C^{15}_{12}}=\frac{\frac{10*9}{2}*5+10*10+1*10}{\frac{15*14*13}{3*2}}=\frac{5*(45+20+2)}{5*91}=\frac{67}{91}$$
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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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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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Re: Jury members selection- MGMAT [#permalink]

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21 May 2009, 02:27
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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]

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10 Jun 2009, 15:41
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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]

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09 Aug 2010, 13:02
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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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13 Jan 2011, 09:18
In is difficult to understand the question stem because of incorrect grammatical construction. The right to introduce this question must such “If a jury of 12 people is to be selected randomly from a pool of 15 potential jurors, and the jurors’ pool consists of 2/3 men and 1/3 women, what is the probability that at least 2/3 of the jury will comprise of men?”

If the question stem were in the construction I have formulated above everybody would catch the point better. Frankly speaking it was difficult for me to recognize the pattern and to comprehend conditions of this question. I would request gmatraider to be more precise and orderly in representation of his/her questions.
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Re: Jury members selection- MGMAT [#permalink]

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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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05 Apr 2011, 19:42
Total 2/3 men = 10 Men

Total women = 5

So the Probablity that there will be at least 2/3 men in 12 members = 2/3 * 12 = 8

8M4W, 9M3W, 10M2W

= 10C8 * 5C4 + 10C9 * 5C3 + 10C10 * 5C2

= 10!/(2!*8!) * 5 + 10 * 5!/(3!2!) + 1 * 5!/(3!2!)

= 10 * 9/2 * 5 + 10 * 5*4/2 + 5*4/2

= 45 * 5 + 100 + 10

= 225 + 110 = 335

Total ways = 15C12 = (15 * 14 * 13)/(3 * 2)

= 5 * 7 * 13

So Prob = 335/13 * 35 = 67/91

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24 Apr 2011, 19:03
pool consists of 2/3 men 1/3 women

= > men = (2/3)*15 = 10
women = (1/3)*15 =5

total ways of choosing 12 people out of 15 = 15c12

favorable outcomes = 10c8*5c4 + 10c9*5c3 + 10c10*5c2 ( its not possible to chose 11 men , 12 men as we only have 10 men availabity)

= (10c8*5c4 + 10c9*5c3 + 10c10*5c2)/15c12 = 67/91

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Re: 700 plus level question [#permalink]

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03 Aug 2011, 06:45
In the group, there are 10 men & 5 women. So, a jury with atleast 2/3 men (~8 men) can be calculated as follows:

Desired jury could have either 8, 9 or 10 men and rest will be women accordingly.

We need to calculate: (10c8 * 5c4) + (10c9 * 5c3) + (10c10 * 5c2)

(10c8 * 5c4) is explained below:

4 women can be chosen in 5c4 ways = 5 ways
8 men can be chosen in 10c8 ways = 45 ways

Total ways to choose the desired jury of 12 = 5*45 = 225

similarly, we can solve rest and we will get 225+100+10 ways = 335 ways

Total ways to choose 12 from 15 is 15c12 = 455

Probability = 335/455 = 67/91.
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Re: 700 plus level question [#permalink]

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04 Aug 2011, 03:05
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ruturaj 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?
24/91
45/91
2/3
67/91
84/91
Please explain the steps to be followed?

Another way to approach it:

Total jurors - 15
Men = 2/3 = 10
Women = 1/3 = 5

Jurors to be selected = 12
Men - atleast 2/3 = at least 8
Women - at most 1/3 = at most 4

The number of women selected can be 5 or less than 5. Less than 5 means at most 4. So if I find the probability of selecting 5 women and subtract it from 1, we will get the probability of selecting at most 4 women (the required probability)

Total number of ways of selecting the jurors = 15C12 = 15*14*13/3! = 455

Number of ways of selecting 5 women jurors (and 7 men) = 5C5 * 10C7 = 10*9*8/3! = 120

Probability of selecting 5 women jurors = 120/455 = 24/91
Probability of selecting at most 4 women jurors = 1 - 24/91 = 67/91
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Re: Jury members selection- MGMAT [#permalink]

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

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

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03 Aug 2014, 17:19
Hi,

Can someone explain why my method below is incorrect?

I tried to approach it with the "1-x" method and therefore, I did a very simple calc of:

1-(5/12) where the (5/12) is the probability of choosing 5 women, therefore, only 7 men. Why is that incorrect? Why do we need to apply the combination method here and not simple probability?
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Re: If a jury of 12 people is to be selected randomly from a [#permalink]

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03 Aug 2014, 20:33
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@ russ9 : your method is incorrect,

if there are 7 men and 5 women, and we are asked to chose 1 person, then the probability of that 'one person' being a woman is 5/12. this is simple probability...

in this case we are not choosing one person, we are actually choosing 12 people, from a pool of 15 people, comprising of both men and women. So what we are trying to find is when selecting 12 people, how many combinations of those 12 people would contain 5 women? thats what we trying to find out.. that why we use combinations and not simple probability.

Hope its clear
Re: If a jury of 12 people is to be selected randomly from a   [#permalink] 03 Aug 2014, 20:33

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