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. 5/91
c. 2/3
d. 67/91
e. 84/91

barcebal wrote:NOTE: MY QUESTION IS NOT HOW TO SOLVE THIS QUESTION PER SE. I NEED SOMETHING CLARIFIED BUT AM USING THIS QUESTION AS REFERENCE. SEE BELOW.

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. 5/91
c. 2/3
d. 67/91
e. 84/91

So I get that you can solve by doing 1 - probability of 5 women on the jury.
So to do that you'd do ( 5C5 women * 10C7 men ) divided by total possible juries (15C12).
From there you get 1 - ((5!/5!0!)*(10!/7!3!))/(15!/12!3!) or (10!12!3!/7!3!15!)

But I'm wondering why I can't do THIS:
Find probability of getting ONE all-women on jury: 5/15 * 4/14 * 3/13 * 2/12 * 1/11 * 10/10....which becomes (5!10!/15!)
and then multiplying that probability by the total number of juries where this could occur which is essentially saying how can I have 10 men, choose 7. (10!/7!3!)

But this give me a much different answer:

So to recap, the right way gives me 1 - 10!12!3!/7!3!15!
and the wrong way gives me 1 - 10!10!5!/7!3!15!

So what am I misinterpreting on the WRONG way?

kullayappayenugula wrote:Jim can you please suggest an approach to solve the above problem in 2 min??