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Math Help - Drawing cards without replacement.

  1. #1
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    Drawing cards without replacement.

    Two cards are drawn without replacement from a shuffled deck of 52 cards. What is the probability that the second card is a king?
    Solutions showing step-by-step is very much appreciated.
    Thank you in advance .
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  2. #2
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    Hello, flywithme!

    Two cards are drawn without replacement from a shuffled deck of 52 cards.
    What is the probability that the second card is a King?
    I can solve this The Long Way, but why bother?

    P(\text{2nd is King}) \;=\;\frac{4}{52} \:=\:\frac{1}{13}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Do we care what the first card is? . . . No!


    Suppose they asked for the probability that the 17th card is a King.
    Do we need to know if any of the first 16 cards are Kings? . . . No.


    Spread the cards face down on the table.
    . . Point to any card: "Is it a King?"
    The probability will be: . \frac{4}{52} \,=\,\frac{1}{13}

    Get it?

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  3. #3
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    Quote Originally Posted by flywithme View Post
    Two cards are drawn without replacement from a shuffled deck of 52 cards. What is the probability that the second card is a king? Solutions showing step-by-step is very much appreciated.
    Why don't you show us some effort on your part?

    EDIT: There is always a spoil-sport who must show that he can do the question.
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  4. #4
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    Thank you very much Soroban, I never tackled probability with abstract thoughts.
    @Plato, how can I show effort on my part? I struggle hard with probability and I can't help others with other types of math because I am only mediocre.
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  5. #5
    MHF Contributor matheagle's Avatar
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    I always talk about this in class.
    I usually ask what is the probability we select the ace of spades on the second, fifth... pick?
    It's always 1/52.
    BUT then I prove it....

    P(King on second pick)=P(KING, KING)+P(not a KING, KING)

    = \left({4\over 52}\right)\left({3\over 51}\right) +\left({48\over 52}\right)\left({4\over 51}\right)

    = \left({4\over 52}\right)\left[{3\over 51}+{48\over 51}\right]

     = {4\over 52}= {1\over 13}
    Last edited by matheagle; January 24th 2010 at 07:49 AM.
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Plato View Post
    EDIT: There is always a spoil-sport who must show that he can do the question.
    I'm sorry, but I really can't help but laugh...
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