Another way to approach it:

Total jurors - 15
Men = 2/3 = 10
Women = 1/3 = 5

Jurors to be selected = 12
Men - atleast 2/3 = at least 8
Women - at most 1/3 = at most 4

The number of women selected can be 5 or less than 5. Less than 5 means at most 4. So if I find the probability of selecting 5 women and subtract it from 1, we will get the probability of selecting at most 4 women (the required probability)

Total number of ways of selecting the jurors = 15C12 = 15*14*13/3! = 455

Number of ways of selecting 5 women jurors (and 7 men) = 5C5 * 10C7 = 10*9*8/3! = 120

Probability of selecting 5 women jurors = 120/455 = 24/91
Probability of selecting at most 4 women jurors = 1 - 24/91 = 67/91
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clear answer, thanks - one question though, the numerator is 455-120; is that 120 because you have multiplied by the 1? i.e. if there were two ways, would it have been 120*2 leading to 455-240?

D

\(P=\frac{C^{10}_{8}*C^{5}_{4}+C^{10}_{9}*C^{5}_{3}+C^{10}_{10}*C^{5}_{2}}{C^{15}_{12}}=\frac{\frac{10*9}{2}*5+10*10+1*10}{\frac{15*14*13}{3*2}}=\frac{5*(45+20+2)}{5*91}=\frac{67}{91}\)
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IMO = D

Out of 15 juries, there are 10 men and 5 women.

1) 8 men, 4 women = 225 ways
2) 9 men, 3 women = 100 ways
3) 10 men, 2 women = 10 ways

The total number of ways to select is 335 ways out of 455 (15C12 - the number of possible ways) or 67 / 91

There can be below scenarios.
a). 8M + 4W = 10C8 5C4 = 225 ways.
b). 9M + 3W = 10C9 5C3 = 100
c). 10M + 2W = 10C10 5C2 = 10
So total selections possible = a+b+c = 335
Total ways to select 12 = 15C12 = 455

Probability = 335/455 = 67/91.

In is difficult to understand the question stem because of incorrect grammatical construction. The right to introduce this question must such “If a jury of 12 people is to be selected randomly from a pool of 15 potential jurors, and the jurors’ pool consists of 2/3 men and 1/3 women, what is the probability that at least 2/3 of the jury will comprise of men?”

If the question stem were in the construction I have formulated above everybody would catch the point better. Frankly speaking it was difficult for me to recognize the pattern and to comprehend conditions of this question. I would request gmatraider to be more precise and orderly in representation of his/her questions.

Total 2/3 men = 10 Men

Total women = 5

So the Probablity that there will be at least 2/3 men in 12 members = 2/3 * 12 = 8


8M4W, 9M3W, 10M2W

= 10C8 * 5C4 + 10C9 * 5C3 + 10C10 * 5C2

= 10!/(2!*8!) * 5 + 10 * 5!/(3!2!) + 1 * 5!/(3!2!)

= 10 * 9/2 * 5 + 10 * 5*4/2 + 5*4/2

= 45 * 5 + 100 + 10

= 225 + 110 = 335

Total ways = 15C12 = (15 * 14 * 13)/(3 * 2)


= 5 * 7 * 13

So Prob = 335/13 * 35 = 67/91

Answer - D
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In the group, there are 10 men & 5 women. So, a jury with atleast 2/3 men (~8 men) can be calculated as follows:

Desired jury could have either 8, 9 or 10 men and rest will be women accordingly.

We need to calculate: (10c8 * 5c4) + (10c9 * 5c3) + (10c10 * 5c2)

(10c8 * 5c4) is explained below:

4 women can be chosen in 5c4 ways = 5 ways
8 men can be chosen in 10c8 ways = 45 ways

Total ways to choose the desired jury of 12 = 5*45 = 225

similarly, we can solve rest and we will get 225+100+10 ways = 335 ways

Total ways to choose 12 from 15 is 15c12 = 455

Probability = 335/455 = 67/91.

Hi,

Can someone explain why my method below is incorrect?

I tried to approach it with the "1-x" method and therefore, I did a very simple calc of:

1-(5/12) where the (5/12) is the probability of choosing 5 women, therefore, only 7 men. Why is that incorrect? Why do we need to apply the combination method here and not simple probability?

Hi

My doubt is:

For choosing 7 Men and 5 Women out of a group of 15 people (10M and 5W)...

What about this:

P ( of getting 7 Men and 5 Women) = \(\frac{10*9*8*7*6*5*4}{15*14*13*12*11*10*9}* \frac{5*4*3*2*1}{8*7*6*5*4}\)

I know it is wrong.. but what is the fault in this logic?

Please clarify my faulty logic..

How can we solve this problem using the probability route -- and not the combinations way..?