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Permutations

by cssilverman » Mon Apr 11, 2011 8:30 pm
I am still confused about how to compute permutations. Here is an example problem that is causing me trouble:

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

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by shovan85 » Mon Apr 11, 2011 8:57 pm
cssilverman wrote:I am still confused about how to compute permutations. Here is an example problem that is causing me trouble:

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
Need to form a group of 12 people.
This can be done in C(15,12) = 455

2/3 of the 15 are Men = 10 men
1/3 of the 15 are women = 5 women

in the formed group 2/3 at least are needed to be men,
so at least 8 men needed.

C(10,8) * C(5,4) + C(10,9) * C(5,3) + C(10,10) * C(5,2)
= 225 + 100 + 10
= 335 ways

The required probabilty = 335/455 = 67/91
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by [email protected] » Mon Apr 11, 2011 9:14 pm
Hi!

This isn't actually a permutations question, since order doesn't matter. It can be solved with either combinations or probability.

Here's another thread discussing the question: https://www.beatthegmat.com/i-suck-at-pr ... t8927.html
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