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Math Help - Probability

  1. #1
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    Question Probability

    Hi folks!

    I've this one to solve, but I just can't figure it. It is probably quite simple however.

    A 4 person leadership committee is randomly chosen from a group of 24 candidates. Ten of the candidates are men, and 14 are women. Find the probability that:

    A. The committe is all male or all female

    B. The commitee has at least 1 man or at least 1 woman

    I know this is using mutually excluisive/inclusive events, and possibly a permutation/combination thing somewhere.

    Thanks in advance folks!
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  2. #2
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    so we start out by seeing how many ways we can select 4 out of 24, which is:

     \binom{24}{4} = 10 ,626

    so for all male we would have:

     P(M=4) = \frac{\binom{10}{4} \binom{14}{0}}{\binom{24}{4}} = \frac{(210)(1)}{10,626} = \frac{5}{253}

    for all females it would be:
     P(F=4) = \frac{\binom{10}{0} \binom{14}{4}}{\binom{24}{4}} = \frac{(1)(1001)}{10,626} = \frac{13}{138}

    at least 1 male is:

     P(M \geq 1) = P(M=1) \cup P(M=2) \cup P(M=3) \cup P(M=4) \Rightarrow 1-P(M=0)

     1-\frac{(1)(1001)}{10,626} = \frac{125}{138}


    at least 1 female:

     P(F \geq 1) = P(F=1) \cup P(F=2) \cup P(F=3) \cup P(F=4) \Rightarrow 1-P(F=0)

     1-\frac{(210)(1)}{10,626} = \frac{248}{253}
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  3. #3
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    Right so I get everything else but for the "at least one male or at least one female" bit--- how would you solve that?
    I suppose one would have to find the probability of both happening.
    Would that be had by multiplying 125/138 by 248/253 ?

    so P(at least one man or at least one woman) = 125/138 + 248/253 - 31000/34914 = approx. 0.9981

    Right?
    Last edited by scotland4ever; May 5th 2008 at 04:34 PM.
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  4. #4
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    Quote Originally Posted by scotland4ever View Post
    I'm not quite understanding...
    could you explain it more??

    thanks very much.
    It would be easier to explain it more if you pointed out the bits you don't understand.
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  5. #5
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    sorry. I have edited my post accordingly.
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