Problem Solving

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?

What you are calculating is not the probabilities you are claiming. The probability of a 5 woman jury is (10 Choose 7) / (15 choose 12). You are mixing and matching the equations. Please let me know if this doesn't make sense.

I get how to do it the way you explained, but most probabilities have various ways to solve. I'm confused why the following method won't work.

Why can't I say the probability of getting a five-women (don't care which men are on it at this point) jury is:

5/15*4/14*3/13*2/12*1/11*(don't need to multiply the men into it because if you get the 5 women, you will get all the men).

Then to find how many different all-women juries you would multiply the result by 10C7 (10!/7!3!)

Obviously this doesn't work. I don't get why...what am I calculating instead?

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

my solution is some thing like the below but it had lot of calcutations to be performed.

total no. of jury - 15
jury to be selected - 12

out of jury of 15 members - 10 are men and 5 are women.
since the question says the jury should consist 2/3 men we need to have atleast 8 men and no.of women can vary.

i.e
8 men 4 women
9 men 3 women
10 men 2 women


=> 12c8*5c4+12c9*5c3+12c10*5c2/15c12

now I am sure that the above calcutaion will take more than 5 min . Please help.

i don't know how you can imply 5 women for the conditions given in this question.

Given Pool: Men=10, Women=5
Required Jury: Men=at least 8 (possible 8,9,10, AND NOT 11,12 as we are limited to ONLY 10 men from Pool) and Women=possible 4,3,2,1,0

as you see women can be at most 4 not 5. This is important, and I don't know if you have this question solved and answered to verify your answer, but I will solve it now.

In your *wondered* solution you make the same mistake of 5 women instead of considering 4 women at most. Anyways i'm directing myself to solve this >>

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

The required Probability (8M&4W + 9M&3W + 10M&2W) -> 10C8*5C4/15C12 + 10C9*5C3/15C12 + 10C10*5C2/15C12 = 45*5/455 + 10*10/455 + 1*10/455 = 335/455 = 67*5/91*5 = 67/91

Correct answer is d

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??

my sol. took me less than 2 min-s and i spent time more describing process in the thread, otherwise combo wraps+quick calc=<2 min-s

I still don't see why you can't calculate the odds/probability of getting an all-women jury

5/15 * 4/14 * 3/13 * 2/12 * 1/11

And then finding number of different combinations of 7 men from a field of 10.