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by bryan88 » Sun May 06, 2012 10:03 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

What is the 1-x method to this?
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by aneesh.kg » Sun May 06, 2012 10:25 am
A team of 12 has to be selected from a pool of 10 men and 5 women.

The 1-x method:
The only unfavourable case is when in the team of 12, we have 7 Men and 5 Women.
So, we will subtract the probability of that from 1.

The required probability is:
1 - [P(7M,5W)]
= 1 - [10C7*5C5/15C12]
= 1 - (120/455)
= 335/455
= [spoiler]67/91 (Option D)[/spoiler]

Alternatively:
We will add up the probability of all the favourable outcomes.

The required probability is:
P(8M,4W) + P(9M,3W) + P(10M, 2W)
= 10C8*5C4 / 15C12 + 10C9*5C3 / 15C12 + 10C10*5C2 / 15C12
= (45*5 + 10*10 + 1*10)/(5*7*13)
= [spoiler]67/91 (Option D)[/spoiler]

Same Answer.
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by bryan88 » Sun May 06, 2012 10:40 am
A case of 6 men and 6 women wont exists as the number of women is capped at 5.
Tricky question under time constraints.
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by Anurag@Gurome » Sun May 06, 2012 7:55 pm
bryan88 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

What is the 1-x method to this?
The following combinations are possible: (8M, 4W), (9M, 3W), (10M, 2W)
Total number of ways of selecting a jury of 12 people from 15 people = 15C12 = (15 * 14 * 13)/3 * 2) = 5 * 7 * 13
Required Probability = (10C8 * 5C4) + (10C9 * 5C3) + (10C10 * 5C2)/(5 * 7 * 13)
= [(5 * 9 * 5) + (10 * 5 * 2) + (5 * 2)]/(5 * 7 * 13)
= [(9 * 5) + (10 * 2) + 2]/(7 * 13)
= [spoiler]67/91[/spoiler]

The correct answer is D.
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