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Jury selection probability

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Jury selection probability

by gmattesttaker2 » Sun Oct 13, 2013 5:43 pm
Hello,

This problem has been asked before on these formums:

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?

OA: [spoiler]67/91[/spoiler]


I was just wondering if the following method is correct:

Total Probability = P (8M and 4W) + P (9M and 3W) + P (10 M and 2 W)

P ( 8M and 4W ) = P (8M). P (4W)
P ( 8M ) = 8/10.7/9.6/8.5/7.4/6.3/5.2/4.1/3 = 1/45
P ( 4W ) = 4/5.3/4.2/3.1/2 = 1/5

So, P ( 8M and 4W ) = P (8M). P (4W) = 1/45 . 1/5 = 1/225


P ( 9M and 3W ) = P ( 9M ) . P ( 3W )
P ( 9M ) = 9/10.8/9.7/8.6/7.5/6.4/5.3/4.2/3.1/2 = 1/10
p ( 3W ) = 3/5.2/4.1/3 = 2/20

So, P ( 9M and 3W ) = P ( 9M ) . P ( 3W ) = 1/10.2/20 = 1/100


P ( 10M and 2W ) = P ( 10M ). P ( 2W )
P ( 10M ) = 1
P ( 2W ) = 2/5.1/4 = 1/10

So, P ( 10M and 2W ) = P ( 10M ). P ( 2W ) = 1.1/10 = 1/10

Total Probability = P (8M and 4W) + P (9M and 3W) + P (10 M and 2 W)
= 1/225 + 1/100 + 1/10

However, I think I am going wrong somewhere since I don't get the correct answer. Can you please help?

Best Regards,
Sri
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by rakeshd347 » Sun Oct 13, 2013 7:34 pm
gmattesttaker2 wrote:Hello,

This problem has been asked before on these formums:

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?

OA: [spoiler]67/91[/spoiler]


I was just wondering if the following method is correct:

Total Probability = P (8M and 4W) + P (9M and 3W) + P (10 M and 2 W)

P ( 8M and 4W ) = P (8M). P (4W)
P ( 8M ) = 8/10.7/9.6/8.5/7.4/6.3/5.2/4.1/3 = 1/45
P ( 4W ) = 4/5.3/4.2/3.1/2 = 1/5

So, P ( 8M and 4W ) = P (8M). P (4W) = 1/45 . 1/5 = 1/225


P ( 9M and 3W ) = P ( 9M ) . P ( 3W )
P ( 9M ) = 9/10.8/9.7/8.6/7.5/6.4/5.3/4.2/3.1/2 = 1/10
p ( 3W ) = 3/5.2/4.1/3 = 2/20

So, P ( 9M and 3W ) = P ( 9M ) . P ( 3W ) = 1/10.2/20 = 1/100


P ( 10M and 2W ) = P ( 10M ). P ( 2W )
P ( 10M ) = 1
P ( 2W ) = 2/5.1/4 = 1/10

So, P ( 10M and 2W ) = P ( 10M ). P ( 2W ) = 1.1/10 = 1/10

Total Probability = P (8M and 4W) + P (9M and 3W) + P (10 M and 2 W)
= 1/225 + 1/100 + 1/10

However, I think I am going wrong somewhere since I don't get the correct answer. Can you please help?

Best Regards,
Sri
Hi Sri,

In the at least questions it is always good to calculate the probability of not having that number and then do 1-not having.

Here we have:
Number of men=10
Number of women=5
Jury to be chosen=12

So total number of ways choosing 12 people out of 15= 15C12=455
Now consider one case that we don't have at least 8 (2/3 men) in the jury. Then we have to select all the 5 women....there is only one way to choose the 5 women out of 5.
But to chose 7 men out of 10 we have 10C7=120 ways....So there are 120 ways in which you can choose 7 men out of 10 mens.

Now the probability that at least 2/3 men won't be in the jury=total number of ways to chose 5 women & 7 men/total number of ways=120/455=24/91

Now here we are asked total number of ways in which we have at least 2/3 (8 men) in the jury.
So it will be 1-probability of not having at least 2/3 men=1-24/91=67/91

So the final answer is [spoiler]67/91[/spoiler]
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by ceilidh.erickson » Tue Oct 15, 2013 5:07 pm
Rakesh's explanation is great. Just on a more general note, whenever you see the language of "AT LEAST" on a GMAT problem, there's probably only one scenario that wouldn't work (I've never seen a real GMAT problem where there was more than one that wouldn't work). Always think about the opposite scenario, and subtract that from 1. That will be a much quicker way to answer the question.

Examples from OG13:
PS 193
DS D45
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
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by gmattesttaker2 » Thu Oct 17, 2013 10:29 pm
Hi,

I was trying to use the 1 - x approach and solve as follows:

P ( At least 8 Men ) = 1 - P ( less than 8 M )
= 1 - P ( 7 Men and 5 Women )


P ( 7 Men ) = 10/15 . 9/14 . 8/13 . 7/12 . 6/11 . 5/10 . 4/9
= 8/13 . 1/11 . 1/3


P ( 5 Women ) = 5/15 . 4/14 . 3/13 . 2/12 . 1/11
= 1/21 . 1/13 . 1/11


I was wondering if this approach is correct. I think I am going wrong somewhere though since the answer doesn't seem right. But I mainly wanted to find out if I can use the above method to solve these kinds of problems. Thanks for your help - Sri
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by ceilidh.erickson » Fri Oct 18, 2013 6:08 am
gmattesttaker2 wrote:Hi,

I was trying to use the 1 - x approach and solve as follows:

P ( At least 8 Men ) = 1 - P ( less than 8 M )
= 1 - P ( 7 Men and 5 Women )


P ( 7 Men ) = 10/15 . 9/14 . 8/13 . 7/12 . 6/11 . 5/10 . 4/9
= 8/13 . 1/11 . 1/3


P ( 5 Women ) = 5/15 . 4/14 . 3/13 . 2/12 . 1/11
= 1/21 . 1/13 . 1/11


I was wondering if this approach is correct. I think I am going wrong somewhere though since the answer doesn't seem right. But I mainly wanted to find out if I can use the above method to solve these kinds of problems. Thanks for your help - Sri
The problem is that you're not answering the right question here.

What you set up is the probability of choosing a man first, then a man second, etc, then choosing a woman 8th, 9th, and 10th. That's one way to choose 7 men and 5 women, but it's not the only way - they don't have to be in that specific order. That's why the answer you found was much smaller than it should have been.

A better way to solve this problem (because men and women can be chosen in any order) is to think of the entire scenario, and not individual picks: DESIRED OUTCOME / TOTAL POSSIBLE OUTCOMES.

DESIRED OUTCOME: choose 7 out of the 10 men, and 5 out of the 5 women.
choose 7 out of 10 (where order doesn't matter): 10!/7!3! = 120
choose 5 out of 5 women: there's only 1 way to choose all 5 women

TOTAL POSSIBLE OUTCOMES: all of the ways to choose 12 people out of 15
choose 12 out of 15 (where order doesn't matter): 15!/12!3! = 455

DESIRED OUTCOME / TOTAL POSSIBLE: 120/455 = 24/91

Now use 1 - x: 1 - 24/91 = 67/91
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
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