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Probability clarification

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barcebal Rising GMAT Star Default Avatar
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Probability clarification Post Tue Feb 07, 2012 6:31 pm
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  • Lap #[LAPCOUNT] ([LAPTIME])
    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?

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    Jim@StratusPrep MBA Admissions Consultant
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    Post Tue Feb 07, 2012 10:48 pm
    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.

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    barcebal Rising GMAT Star Default Avatar
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    Post Tue Feb 07, 2012 10:55 pm
    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?

    kullayappayenugula Rising GMAT Star Default Avatar
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    Post Tue Feb 07, 2012 11:03 pm
    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 Sad. Please help.



    Last edited by kullayappayenugula on Tue Feb 07, 2012 11:06 pm; edited 1 time in total

    pemdas GMAT Titan Default Avatar
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    Post Tue Feb 07, 2012 11:05 pm
    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 >>
    Quote:
    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?

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    pemdas GMAT Titan Default Avatar
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    Post Tue Feb 07, 2012 11:08 pm
    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

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    barcebal Rising GMAT Star Default Avatar
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    Post Wed Feb 15, 2012 1:12 pm
    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.

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