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Math Help - probability question

  1. #1
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    probability question

    A college delegation is made up of 6 basketball players, 5 volleyball players, 4 netball players, 3 badminton players and 2 tennis players. A subcommittee of four people is to be formed from these 20 athletes.

    Find the probability that the subcommittee contains exactly one basketball player or exactly one volleyball player (or both).

    My method:

    let A be the event that committee has one basketball player
    let B be event that committee has one volleyball player

    so we want P (A U B)= P(A) + P(B) - P (A n B)

    (n is intersect)

    P(A) = 6C1 x 14C3/20C4
    P(B) = 5C1 x 15C3/20C4
    P(A n B) = 6C1 x 5C1 x9C2/20C4

    my answer is 0.697 but the answer given is 0.414 can someone tell me where i went wrong? thanks~
    Last edited by thesocialnetwork; April 29th 2012 at 05:11 AM.
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  2. #2
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    Re: probability question

    Quote Originally Posted by thesocialnetwork View Post
    A college delegation is made up of 6 basketball players, 5 volleyball players, 4 netball players, 3 badminton players and 2 tennis players. A subcommittee of four people is to be formed from these 20 athletes.
    Find the probability that the subcommittee contains exactly one basketball player or exactly one volleyball player (or both).
    my answer is 0.697 but the answer given is 0.414 can someone tell me where i went wrong? thanks~
    The wording is quite awkward. The following gives the suggested answer:
    \dfrac{\dbinom{6}{1}\dbinom{9}{3}+\dbinom{5}{1} \dbinom{9}{3}+\dbinom{6}{1}\dbinom{5}{1}\dbinom{9}  {2}}{\dbinom{20}{4}}.

    Can you tell us why?
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  3. #3
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    Re: probability question

    yes, you take the 3 cases separately: (1) where there is exactly one basketball player and no volleyball player, (2) where there is exactly one volleyball player and no basketball player and (3) where there is exactly one basketball player and exactly one volleyball player, and then you divide the total number of outcomes in these 3 events by the total possible number of outcomes.

    Technically though, isn't the method I first used correct? Where I add up P(A) and P(B) and subtract the intersect? Did I make a calculation error? I would've thought I answered the question properly.
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  4. #4
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    Re: probability question

    I think that what is happening is that your event A could include exactly one of each and so could your event B.
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  5. #5
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    Re: probability question

    Quote Originally Posted by thesocialnetwork View Post
    Technically though, isn't the method I first used correct? Where I add up P(A) and P(B) and subtract the intersect? Did I make a calculation error? I would've thought I answered the question properly.
    Your event A includes the case of exactly one baseball player and possibly three volleyball players. We don't allow that. We want exactly one baseball player, or exactly one volleyball player or exactly one of each. Those are exclusive or's.
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  6. #6
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    Re: probability question

    yes i see - the overlap is greater than i thought and my P (A n B) doesn't account for it. thank you very much! It's appreciated.
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