Two cards are drawn at random with replacement from a deck of 52 cards. What is the probability that both card

Two cards are drawn at random with replacement from a deck of 52 cards. What is the probability that both cards are aces?

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  • Como
    Lv 7
    1 decade ago
    Best Answer

    P (1st card is an ace) = 4 / 52 = 1 / 13

    P (2nd card is an ace) = 4 / 52 = 1 / 13

    P (both aces) = (1/13) (1/13) = 1 / 169

  • 1 decade ago

    If this is a truly random drawing, and I assume they are drawn one at a time, then the probability is defined by the likelihood of getting an ace each of the two times.

    Since there are 4 aces in a 52 card deck, the probability of drawing one is 4/52, or 1/13. Drawing two results in the probability of 4/52 * 4/52, or 1/(13*13) = 1/169

  • 4 years ago

    Your chance of picking an ace out of a standard pack of cards is 4/52, if you did get an ace and you did not replace it your chance of picking another ace is 3/51. Now multiply the two probabilities into each other: 4/52 x 3/51 Remember when multiplying fractions you can simplify either vertically or diagonally, NEVER horizontally, so simplify the 4 and 52: 1/13 x 3/51 Now simplify the 3 and 51: 1/13 x 1/17 = 1/221

  • 1 decade ago

    The probability of drawing an ace is 4/52 = 1/13.

    The probability of drawing two aces = 1/13 * 1/13 = 1/169.

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  • 1 decade ago

    The probability of both is 1:169.

    There are 4 aces in a deck. To calculate this, multiply 4/52 by 4/52 and you get 16:2704. If you reduce this, you get 1:169.

  • Merlyn
    Lv 7
    1 decade ago

    Let X be the number of Ace's drawn. X has the binomial distribution with n = 2 trials and success probability of 1/13.

    In general, if X has the binomial distribution with n trials and a success probability of p then

    P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)

    for values of x = 0, 1, 2, ..., n

    P[X = x] = 0 for any other value of x.

    for this question you have to find P(X = 2)

    plug in the values into the probability mass function above and you'll find:

    P(X = 2) = 0.00591716

  • 1 decade ago

    To draw one ace is 1/13. This is true both times because there is replacement.

    For both events to happen you multiply the prob so

    1/13 X1/13 =1/169

  • 1 decade ago

    221 to 1

    4/52 * 3/51

  • Anonymous
    1 decade ago

    1/169

    (1/13)^2

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