A five-card poker hand is dealt at random from a standard 52-card deck. Note the total number of possible hands is C(52,5)=2,598,960. Find t

Question

A five-card poker hand is dealt at random from a standard 52-card deck. Note the total number of possible hands is C(52,5)=2,598,960. Find the probabilities of the following scenarios: (a) What is the probability that the hand contains exactly one ace?

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Parker 1 month 2021-10-19T23:19:14+00:00 1 Answer 0 views 0

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    2021-10-19T23:20:46+00:00

    Answer:

    (a) Probability = 0.29947

    Step-by-step explanation:

    The probability of the hand containing exactly one ace would be:

    Number of ways this can happen = 4C1 * 48C4      (using combinations)

    Number of ways this can happen = 4 * 194580

    Number of ways this can happen = 778,320

    Total number possible hands = 2,598,960 (as stated in question)

    Total probability of exactly one ace = Number of ways to get an ace / total number of ways

    Total probability = 778320 / 2598960 = 0.29947

    Thus, the probability of the hand containing exactly one ace will be 0.2994

    Another way to solve this:

    Probability of one ace and 5 other cards = \frac{4}{52}*\frac{48}{51}*\frac{47}{50}*\frac{46}{49}*\frac{45}{48} = 0.059894

    Number of ways to arrange 1 ace and 4 other cards = 5

    Total probability = 0.0598 * 5 = 0.29947

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45:7+7-4:2-5:5*4+35:2 =? ( )