Probability Question (Easy)?

--How many different 5-cards can be dealt that contains no aces, from a regular 52-card deck?

--How many different 5-card can be dealt that contains "a" aces (for a = 0 to 4), from a regular 52-card deck?

Thank you for helping.

Update:

Can you show me some steps? My questions are asking for how many ways, not for the probability.

2 Answers

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  • Chris
    Lv 7
    10 years ago
    Favorite Answer

    1. This is the number of combinations of 48 cards, 5 at a time, because aces are excluded from a 52-card deck. You probably know this stuff already, but C(48,5) means the number of combinations of 48 cards, five at a time, and C(n,r) is defined as C(n,r) = n! / (r! * (n-r)!).

    C(4,0) * C(48,5) = 1 * 48!/(43! 5!) = 1,712,304

    2.

    For a = 0, we drew no aces out of the 4 available aces, and 5 cards out of the 48 other cards.

    a = 0: We already answered this one above: 1,712,304

    For a = 1, we drew 1 ace out of the 4 available aces, and 4 cards out of the 48 other cards.

    a = 1: C(4,1) * C(48, 4) = 778,320

    For a = 2, we drew 2 aces out of the 4 available aces, and 3 cards out of the 48 other cards.

    a = 2: C(4,2) * C(48,3) = 103,776

    For a = 3, we drew 3 aces out of the 4 available aces, and 2 cards out of the 48 other cards.

    a = 3: C(4,3) * C(48,2) = 4,512

    For a = 4, we drew all 4 aces out of the 4 available aces, and 1 card out of the 48 other cards.

    a = 4: C(4,4) * C(48,1) = 48

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  • 10 years ago

    First question: Probability = 1/47.

    Second question: 5 X 4C1/52C1.

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