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Math Help - Stuck on a few problems

  1. #1
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    Stuck on a few problems

    What is the Probability that a 5-card hand has at least 3 face cards (K,Q,J) ?
    .07395 is the answer but how?
    12 face cards.
    12 choose 1
    11 choose 1
    10 choose 1
    and I've been using 49 choose 1 and 48 choose 1 for the remaining 2 cards, divided by 2598960 (52 choose 5)




    What is the probability that a 5 card hand has at least one spade?
    .63295 is the answer, didnt know where to start on this one.

    13 cards, 4 suits

    h, d, c, s, h
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  2. #2
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    Quote Originally Posted by shootie300 View Post
    What is the Probability that a 5-card hand has at least 3 face cards (K,Q,J) ?
    .07395 is the answer but how?

    Mr F says: See main post below for some hints.

    12 face cards.
    12 choose 1
    11 choose 1
    10 choose 1
    and I've been using 49 choose 1 and 48 choose 1 for the remaining 2 cards, divided by 2598960 (52 choose 5)




    What is the probability that a 5 card hand has at least one spade?
    .63295 is the answer, didnt know where to start on this one.

    Mr F says: 1 - Pr(no spades).

    13 cards, 4 suits

    h, d, c, s, h
    Number of ways you can have at least 3 face cards  = {12 \choose 3} \, {40 \choose 2} + {12 \choose 4} \, {40 \choose 1} + {12 \choose 5} \, {40 \choose 0} .

    Number of ways of choosing 5 cards {52 \choose 5}.

    Therefore ....
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  3. #3
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    [quote=shootie300;183034]What is the Probability that a 5-card hand has at least 3 face cards (K,Q,J) ?
    .07395 is the answer but how?
    12 face cards.
    12 choose 1
    11 choose 1
    10 choose 1
    and I've been using 49 choose 1 and 48 choose 1 for the remaining 2 cards, divided by 2598960 (52 choose 5)

    Number of ways you can choose at least three face cards
    = (3 face cards)(2 non-face cards) + (4 face cards)(1 non-face cards) +(5 face cards)(0 non-face cards)

    = C{12 \choose 3} \times C{40 \choose 2} + C{12 \choose 4} \times C{40 \choose 1} + C{12 \choose 5} \times C{40 \choose 0}

    = 171600 + 19800 + 792

    = 192192

    Number of ways of choosing 5 cards = C{52 \choose 5}
    = 2598960

    Required probability = \frac {192192}{2598960}

    = 0.7395
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