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Thread: [SOLVED] Urgent help required: Probability

  1. June 22nd 2006, 10:55 PM #1
    stuck_fugu
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    [SOLVED] Urgent help required: Probability

    Two cards are drawn without replacement from a standard deck of 52 cards. What is the probability that the first card is a face card and the second card is a queen?

    I keep getting 7/221

    but the answer says it's 11/663.


    anyone know?
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  2. June 23rd 2006, 01:49 AM #2
    CaptainBlack
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    Quote Originally Posted by stuck_fugu
    Two cards are drawn without replacement from a standard deck of 52 cards. What is the probability that the first card is a face card and the second card is a queen?

    I keep getting 7/221

    but the answer says it's 11/663.


    anyone know?
    <br /> P(1st\ face \land 2nd\ Q)=P(1st\ K \lor N)P(2nd\ Q|1st\ K \lor N) + P(1st\ Q)P(2nd\ Q|1st\ Q)<br />

    (where \land should be read as "and", and \lor should be read as "or".

    Now as there are four of each value of card in a deck and 52 cards:

    <br /> P(1st\ K \lor N)=\frac{8}{52}<br />

    <br /> P(1st\ Q)=\frac{4}{52}<br />

    <br /> P(2nd\ Q|1st\ K \lor N)=\frac{4}{51}<br />

    <br /> P(2nd\ Q|1st\ Q)=\frac{3}{51}<br /> .

    And I'm sure you can do the final sum for yourself.

    RonL
    Last edited by CaptainBlack; June 23rd 2006 at 02:08 AM.
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  3. June 23rd 2006, 04:24 AM #3
    Soroban
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    Hello, stuck_fugu!

    Two cards are drawn without replacement from a standard deck of 52 cards.
    What is the probability that the first card is a face card and the second card is a queen?

    There are two possible cases:


    [1] The first card is a Queen and the second is a Queen.
    . . . There are 4 choices for the first card; 3 choices for the second.

    . . . P(\text{Q, then Q})\:=\:\frac{4}{52}\cdot\frac{3}{51}\;=\;\frac{3} {663}


    [2] The first card is a Face card (not a Queen) and the second is a Queen.
    . . . There are 8 choices for the first card (4 Jacks, 4 Kings)
    . . . and 4 choices for the second card (4 Queens).

    . . . P(\text{J or K, then Q})\:=\:\frac{8}{52}\cdot\frac{4}{51}\:=\:\frac{8} {663}


    Therefore: . P(\text{Face and Q})\:=\;\frac{3}{663} + \frac{8}{663}\:=\:\frac{11}{663}
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