Probability

[niketdoshi123]

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
d)87/91

[Anurag@Gurome]

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?

Total number of ways to select 12 people from 15 = 15C12 = 15*14*13/(3*2) = 5*91

Number of total men = 2/3 of 15 = 10
Number of total women = 1/3 of 15 = 5

We need at least 2/3 of 12 = 8 men

Number of ways to select 8 men and 4 women = (10C8)*(5C4) = 45*5
Number of ways to select 9 men and 3 women = (10C9)*(5C3) = 10*10
Number of ways to select 10 men and 2 women = (10C10)*(5C2) = 1*10

Hence, total number of ways such that at least 2/3 are men = (45*5 + 10*10 + 10)

Hence, required probability = (45*5 + 10*10 + 10)/(5*91) = (45 + 20 + 2)/91 = 67/91

The correct answer is D.

[GMATGuruNY]

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
d)87/91

P(good outcome) = 1 - P(bad outcome).

Bad outcome:
Since men must constitute at least 2/3 of the 12-member jury, the jury must be composed of at least 8 men, implying a MAXIMUM of 4 women.
The number of women in the jury pool = (1/3)15 = 5.
Thus, there is only one way to exceed the maximum number of women allowed on the jury and get a bad outcome:
5 women, 7 men.

One way to get 5 women and 7 men:
P(WWWWWMMMMMMM) =
= 5/15 * 4/14 * 3/13 * 2/12 * 1/11 * 10/10 * 9/9 * 8/8 * 7/7 * 6/6 * 5/5 * 4/4
= 1/3 * 2/7 * 3/13 * 1/6 * 1/11
= 2/7 * 1/13 * 1/6 * 1/11.

Total possible ways to get 5 women and 7 men:
Any arrangement of the letters WWWWWMMMMMMM will yield a different way to get 5 women and 7 men.
Thus, the probability above must be multiplied by the number of ways to arrange WWWWWMMMMMMM:
2/7 * 1/13 * 1/6 * 1/11 * 12!/5!7! = 24/91.

P(at least 8 men) = 1 - 24/91 = 67/91.

The correct answer is D.

[eagleeye]

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
d)87/91

Mitch and Anurag explained it very well. I ended up doing it in a shorter way, so I felt it was worth mentioning.

We have 15 Jurors, with 10 men and 5 women.
We need to select a pool of 12 where there are 2/3 men at least = 8 Men.

Now since max. number of women in the 12 person jury = 5, in all other cases, we have 8 men always present. So if we take that only case where there are 5 women out, we'll have our answer.
In that case, since we have all 5 women, we only need to select the 12-5 = 7 men from the pool.

Hence the required probability = 1-(10C7)/(15C12)= 1-(10C3)/(15C3)
= 1 - (10.9.8)/(15.14.13) = 1-24/91 = 67/91.

Cheers!