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Math Help - Probability and Statistics

  1. #1
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    Probability and Statistics

    Here's my problem,

    A certain Lottery has 49 numbers, six of which are the winning numbers for a particular game. To play the game each participant choses six numbers. What is the probability of choosing exactly...

    a) six correct numbers

    b) five correct numbers

    c) four correct numbers

    now i know how to find the answer to a) which is 49C6 which equals 13,983,816...so the answer is 1/13,983,816. But what is the formula for finding b. and c....

    i've tried 6C5/49C6 but that's not right, i don't need the answers but does someone know the formula for figuring these out? this would be a huge help. Thanks!
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Yuma, AZ, USA
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    Quote Originally Posted by gwen01 View Post
    Here's my problem,

    A certain Lottery has 49 numbers, six of which are the winning numbers for a particular game. To play the game each participant choses six numbers. What is the probability of choosing exactly...

    a) six correct numbers

    b) five correct numbers

    c) four correct numbers

    now i know how to find the answer to a) which is 49C6 which equals 13,983,816...so the answer is 1/13,983,816. But what is the formula for finding b. and c....

    i've tried 6C5/49C6 but that's not right, i don't need the answers but does someone know the formula for figuring these out? this would be a huge help. Thanks!
    This should work for both b and c. I will do b

    \frac{\binom{5}{5} \binom{44}{1}}{\binom{49}{6}}=\frac{1}{317814}

    There are 44 arrangments of picking 5 correct numbers.
    Last edited by TheEmptySet; June 1st 2008 at 05:18 PM. Reason: Correct answer but wrong reason. :(
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