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  1. #1
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    cards

    A gambler has been dealt with 5 cards: 2 aces, 1 king, 1 five and 1 nine. he discards the 5 and 9 and takes 2 more cards. what is the probability that he ends with a full house?

    my working:

    to get a full house: its either

    3 aces and 2 kings or 2 aces and 3 kings
    P ( 3 aces and 2 kings ) = 2/47 x 3 /46 x 2
    p (2 aces and 3 kings ) = 2/47 x 3 /46

    so i think the answer should be the sum of both...but the answer given was 0.0083..
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  2. #2
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    Quote Originally Posted by alexandrabel90 View Post
    A gambler has been dealt with 5 cards: 2 aces, 1 king, 1 five and 1 nine. he discards the 5 and 9 and takes 2 more cards. what is the probability that he ends with a full house?

    my working:

    to get a full house: its either

    3 aces and 2 kings or 2 aces and 3 kings
    P ( 3 aces and 2 kings ) = 2/47 x 3 /46 x 2
    p (2 aces and 3 kings ) = 2/47 x 3 /46

    so i think the answer should be the sum of both...but the answer given was 0.0083..
    Hi alexandrabel90,

    he will have a full house if he chooses 1 ace from the 2 remaining and 1 king from the 3 remaining..

    The probability of this is \frac{\binom{2}{1}\binom{3}{1}}{\binom{47}{2}}

    he will also have a full house if he choose 2 kings from the remaining 3.

    The probability of this is \frac{\binom{3}{2}}{\binom{47}{2}}

    Adding these probabilities gives the answer
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  3. #3
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    When using the logic you were following, you must break it up into the different possibilities or ways the event can happen.

    In choosing an ace and a king,
    the ace can be chosen first followed by the king,
    or the king can be chosen first followed by the ace.

    Remember, 2 aces and a king have already been taken from the pack,
    so 2 aces and 3 kings remain to choose from among the available 47.

    hence the probability of picking a king and ace is

    \frac{3}{47}\frac{2}{46}+\frac{2}{47}\frac{3}{46}

    The probability of picking 2 more kings instead is

    \frac{3}{47}\frac{2}{46}

    Hence the probability of a full house is \frac{3(6)}{47(46)}
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