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If a jury of 12 people is to be selected randomly from a [#permalink]
04 Feb 2008, 15:40

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Difficulty:

45% (medium)

Question Stats:

52% (02:47) correct
48% (01:57) wrong based on 62 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?

Re: potential jury members [#permalink]
04 Feb 2008, 16:32

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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

Re: potential jury members [#permalink]
06 Feb 2008, 14:35

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

Answer D

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?

Re: potential jury members [#permalink]
06 Feb 2008, 17:54

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

Answer D

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.

Re: Jury members selection- MGMAT [#permalink]
21 May 2009, 01:27

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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.

Re: potential jury members [#permalink]
13 Jan 2011, 08:18

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.

Re: 700 plus level question [#permalink]
04 Aug 2011, 02:05

1

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Expert's post

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
_________________

Re: If a jury of 12 people is to be selected randomly from a [#permalink]
10 Feb 2014, 17:12

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Re: 700 plus level question [#permalink]
18 May 2014, 16:28

VeritasPrepKarishma wrote:

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

Hi Karishma,

I have two questions:

First Question: I'm having a hard time bridging the gap for the 1-x method. Doesn't the "x" part have to encompass EVERYTHING that's not possible, meaning, the polar opposite?

Second Question

I want to see if I understand the logic here:

Probability Method:

Prob of at least 8 men and 4 women + at least 9 men and 3 women + at least 10 men and 2 women (12!/8!4!){ (10/12)(9/11)(8/7)...(3/5)(5/4)(4/3)(3/2)(2/1) } (now since we have 5 women to pick from and 4 tries remaining, would our probability just be one or be more than one? ....this process is tedious and would have to be carried for the remainder of the options. All that being said, is this correct?

I also tried a combination of prob and combinatorics:

15c12(((2/8)^8 * (1/3)^4) + (2/8)^9(1/3)^3...etc)

Last but not least: Straight Probability

(10c8)(5c4)/15c10 + same drill as above.

gmatclubot

Re: 700 plus level question
[#permalink]
18 May 2014, 16:28