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Math Help - Probability of winning exactly one...

  1. #1
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    Probability of winning exactly one...

    Given there were 50 tickets sold, you bought 5, and there are three prizes (1st, 2nd, and 3rd), what is the probability of winning exactly one prize.

    I don't stand how to show the probability of an "exactly one" scenario. I know how to find the probability of "at least" or "none".

    Help is greatly appreciated.
    Last edited by Iceflash234; October 11th 2010 at 02:52 PM. Reason: typos
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  2. #2
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    I see that you have over twenty-five postings.
    By now you should understand that this is not a homework service
    So you need to either post some of your work on a problem or explain what you do not understand about the question.
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  3. #3
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    I don't stand how to show the probability of an "exactly one" scenario. I know how to find the probability of "at least" or "none".
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  4. #4
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    There are \dbinom{3}{1}\dbinom{47}{4} ways to win exactly one prize with five tickets.
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  5. #5
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    Quote Originally Posted by Plato View Post
    There are \dbinom{3}{1}\dbinom{47}{4} ways to win exactly one prize with five tickets.
    So the probability of getting exactly one would be \frac{1}{\dbinom{3}{1}\dbinom{47}{4}} ?


    That makes sense. Thanks! I appreciate it.
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  6. #6
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    Good grief. NO! This totally wrong.
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  7. #7
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    Quote Originally Posted by Plato View Post
    Good grief. NO! This totally wrong.

    Wait what? How? \dbinom{3}{1}\dbinom{47}{4} is the number of ways to win exactly one with 5 tickets.

    Oh wait a minute, I need to figure out the number of total overall possible winning combinations and then divide the number of ways to win by the total number of winning combinations. But regardless of whether I do 50C3 or 50P3, both of which are smaller that the above combinations so they wouldn't work. Now I'm confused.
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  8. #8
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    The answer is \dfrac{\binom{3}{1}\binom{47}{4}}{\binom{50}{5}}
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