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Math Help - Probability combinations help!

  1. #1
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    Probability combinations help!

    Please help,
    What is the probability of having two aces, two kings, and a queen in a five‐card poker hand?

    ,Thanks
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  2. #2
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    Re: Probability combinations help!

    Quote Originally Posted by lucak View Post
    What is the probability of having two aces, two kings, and a queen in a five‐card poker hand?
    There are \binom{4}{2}\binom{4}{2}\binom{4}{1} ways to have two aces, two kings, and a queen in a five‐card poker hand.
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  3. #3
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    Re: Probability combinations help!

    Quote Originally Posted by Plato View Post
    There are \binom{4}{2}\binom{4}{2}\binom{4}{1} ways to have two aces, two kings, and a queen in a five‐card poker hand.
    I understand how you got that that part I get but the answer in the back of the book was (4combination2)(4combination2)(4combination1) that was in the numerator. Then in the denominator it was 52combination5 and the answer was 3 over 216580. I don't understand how they got that answer
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    Re: Probability combinations help!

    Quote Originally Posted by lucak View Post
    I understand how you got that that part I get but the answer in the back of the book was (4combination2)(4combination2)(4combination1) that was in the numerator. Then in the denominator it was 52combination5 and the answer was 3 over 216580. I don't understand how they got that answer
    Do you understand that \binom{N}{k}=\frac{N!}{k!(N-k)!}~?
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    Re: Probability combinations help!

    Quote Originally Posted by Plato View Post
    Do you understand that \binom{N}{k}=\frac{N!}{k!(N-k)!}~?
    Yes I understand that, but how did they get 3 as the top numerator?
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    Re: Probability combinations help!

    Quote Originally Posted by lucak View Post
    Yes I understand that, but how did they get 3 as the top numerator?
    You do not. Your textbook is wrong. \dfrac{\dbinom{4}{2}\dbinom{4}{2}\dbinom{4}{1}} {\dbinom{52}{5}}=\frac{10673}{192629862}}.
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