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Math Help - Condtional probability and deck of cards.

  1. #1
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    If I want to know what the probability is of choosing a queen from a deck of cards given that the previous card was a queen (not replacing the first card in the deck), I think I would use this
    P(A|B) = \frac{P(A \bigcap B)}{P(B)}

    So the probability of B is \frac{1}{52} and the probability of
    but how do I work out P(A \bigcap B) ?

    thanks

    I meant \frac{4}{52}
    Last edited by mr fantastic; May 18th 2009 at 07:29 PM. Reason: Merged posts and tidied up thread
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  2. #2
    Member SENTINEL4's Avatar
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    I will take a shot...If i am wrong please someone correct me.

    Well let B={The first card is a queen} and A={The second card is a queen.} We want to find
    P(A|B) = \frac{P(A \bigcap B)}{P(B)}

    P(A \cap B)=\frac{\left(\begin{array}{cc}4\\1\end{array}\ri  ght)\left(\begin{array}{cc}3\\1\end{array}\right)}  {\left(\begin{array}{cc}52\\2\end{array}\right)}
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  3. #3
    Member SENTINEL4's Avatar
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    Now that i watched it again i think that the solution i gave you is a bit wrong...
    This kind of probabilities confuse me a lot...
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  4. #4
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    If one queen is drawn on the first try how many are left in the 51 remaining cards?
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  5. #5
    Newbie darthfader's Avatar
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    Hi! I'm new to Math, and I'm starting to learn about probabilities on my own. Isn't the problem done this way?

    Since you need the probability that a queen is drawn given that a queen is drawn first then you have

    \dfrac{4}{52}\cdot\dfrac{3}{51} = \frac{1}{221}

    I don't know if this is right.
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  6. #6
    Member SENTINEL4's Avatar
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    Maybe it's very simple and the probability of choosing a queen from a deck of cards given that the previous card was a queen (not replacing the first card in the deck) is just \frac{3}{51} ????
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  7. #7
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    Quote Originally Posted by darthfader View Post
    Hi! I'm new to Math, and I'm starting to learn about probabilities on my own. Isn't the problem done this way?

    Since you need the probability that a queen is drawn given that a queen is drawn first then you have

    \dfrac{4}{52}\cdot\dfrac{3}{51} = \frac{1}{221}

    I don't know if this is right.
    Yes this is right. The odds of getting a pair of queens in Holdem is 1/221.
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  8. #8
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    Quote Originally Posted by Poker Player View Post
    Yes this is right. The odds of getting a pair of queens in Holdem is 1/221.
    No it is not correct,
    The question is: What is the probability of the second card being a queen given that the first card is a queen.
    The correct answer is \frac{3}{51}.
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  9. #9
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    you're both right

    I actually asked 2 questions in that query because I didn't really understand what was asking.
    looking at my question I first asked queen2 given queen1, and then I asked how do I work out queen1 and queen2.
    I thought I was looking for P(A|B), but now I understand better I was actually seeking P(A \bigcap B)
    thanks to all who contributed.
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