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Very tough Probability Question

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by Tame the CAT » Thu Apr 26, 2007 5:29 pm
This is what I did

Numerator: 10!/(8!2!)*5!/4! = 225

Denominator: 15!/(12!*3!) = 455

or 45/91
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by rajesh_ctm » Thu Apr 26, 2007 8:39 pm
I'm getting 67/91. What is the answer?
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Re: Very tough Probability Question

by jayhawk2001 » Thu Apr 26, 2007 9:58 pm
Tame the CAT 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?

A 24/91

B 45/91

C 2/3

D 67/91

E 84/91
Out of 15 Jurors, 10 are men and 5 are women

The possible combinations of Jurors that will yield atleast 2/3 men =

8 men 4 women = 10C8*5C4
9 men 3 women = 10C9*5C3
10 men 2 women=10C10*5C2

Adding it up we get 335

Number of ways to choose 12 Jurors from 15 = 15C12 = 455

Probability = 335 / 455 = 67/91

I'm getting D as well
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by Tame the CAT » Fri Apr 27, 2007 2:31 am
I missed the at least part!! dammit. I'll post the solutions when I get home.
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by RAGS » Fri Apr 27, 2007 4:39 am
i got D
If It Is To Be It Is Up To Me
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by Cybermusings » Sat Apr 28, 2007 1:59 am
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

If the jury pool of 15 has 2/3 men and 1/3 women it has 10 men and 5 women. Now for the jury to have 2/3 men, it should have 2/3*12 = 8 men and 4 women.
Free selection = 15C12 = 15*14*13/3*2*1 = 5*7*13 = 455 selections
Now when there are 8 men + 4 women = 10C8 * 5C4 = 45*5 = 225
When there are 9 men + 3 women = 10C9 * 5C3 = 10*10 = 100
When there are 10 men + 2 women = 10C10 * 5C2 = 1*10 = 10
There can be no more ways since we have selected all the mean from the juror pool.
Hence Probability = 225+100+10 / 455 = 335/455 = 67/91
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