Find the probability of getting 2 Aces, 3 kings, and one 4, and one other card when being dealt 7 cards from a standard 52-card deck.?

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  • 5 years ago
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    First count the number of ways to form the desired hand. For the "other" card, I assume it can't be another ace, king or four.

    From the 4 aces, pick 2 --> C(4,2) = (4 x 3) / 2! = 6 ways

    From the 4 kings, pick 3 --> C(4,3) = 4 ways

    From the 4 fours, pick 1 --> C(4,1) = 4 ways

    From the 40 other cards (4 suits, 10 ranks), pick 1 --> C(40,1) = 40 ways

    Now multiply those together to get the total number of hands:

    6 x 4 x 4 x 40

    = 3,840 ways

    To get the probability, we need to divide by the total ways to pick *any* 7 cards.

    From the 52 cards, pick *any* 7 cards:

    C(52,7) = (52 x 51 x 50 x 49 x 48 x 47 x 46) / 7!

    = 133,784,560 ways

    P(2 aces, 3 kings, 1 four and 1 other card (not ace/king/four))

    = 3,840 / 133,784,560

    That reduces to:

    = 48 / 1,672,307 (or 0.0000287 or about in 1 in 34,800)

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  • Anonymous
    5 years ago

    4C2 for the aces

    times

    4C3 for the kings

    times

    4C1 for the 4

    times

    40 for the other card not an ace, king or 4.

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