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Conditional Probability Question

  1. Feb 3, 2013 #1
    1. The problem statement, all variables and given/known data

    A population has n men and n women. If you where to take 4 people out of the population to form a group what's the probability that there are exactly the same number of men as women in the group .

    2. Relevant equations



    3. The attempt at a solution

    Ok so I thought of this to some extent and am lost.

    The population size is 2n

    (2n)(2n-1)(2n-2)(2n-3) different possible groups if you consider each person of the population to be distinct people without regards to the condition of gender.

    When you take gender into consideration there are

    2*2*2*2 = 2^4 = 16 different combinations when take the gender condition into account and consider each individual to be distinct (that is GGBB is different from BBGG)

    If I consider each man and women to be the same there are a total of 5 different combinations...
    BBBB
    BBBG
    BBGG
    BGGG
    GGGG

    I know the following formula

    P=(A|B) = [itex]\frac{P(A \bigcap B)}{P(B)}[/itex]

    I'm assuming that I need to use this formula. The only problem is that I don't know how to come up with A and B.

    Thanks for any help that nay one can provide.
     
  2. jcsd
  3. Feb 3, 2013 #2

    Dick

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    It's just the number of ways to pick 2 women from n women times the number of ways to pick 2 men from n men divided by the number of ways to pick 4 people from 2n people regardless of gender, isn't it? You seem to be really over complicating this.
     
  4. Feb 3, 2013 #3
    Ya I guess your right about this. I don't know if I can do this notation but can I do

    [itex](N_{K})^{2}[/itex]

    Square N choose K in this notation? I have never seen this before don't see why I can't use this notation.

    I couldn't figure out how to do this in itex code
     
  5. Feb 3, 2013 #4

    Dick

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    I'm not exactly sure what the jist of your question is, but without even using tex, if C(n,k) is the number of ways to choose k objects from n, then what's your answer? And then write it explicitly using a formula for C(n,k).
     
  6. Feb 3, 2013 #5
    my answer is

    [C(n,2)*c(n,2)]/C(n,4)

    I was wondering if you can write

    C(n,2)*C(n,2) as C^2(n,2)?
     
  7. Feb 3, 2013 #6

    Dick

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    I would write it as C(n,2)^2. But that's not terribly important. And the denominator should be C(2n,4), yes? You know C(n,k)=n!/(k!(n-k)!), right? I would guess they want you to use that to simplify [C(n,2)*C(n,2)]/C(2n,4).
     
  8. Feb 4, 2013 #7

    Ray Vickson

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    If your denominator were C(2n,4) then, yes, your answer would be correct. This is just a so-called hypergeometric probability.
     
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