Two cards are drawn one after the other with replacements from a well shuffled pack of cards. What is the probability that the two cards are both kings?

Mathematics

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Answer

1169\frac{1}{169}

Step-by-step solution

4 steps

  1. Identify the total number of cards and the number of kings in a standard deck.

    A standard deck of cards has 52 cards.
    There are 4 kings in a standard deck of cards.

  2. Calculate the probability of drawing a king on the first draw.

    P(King on 1st draw)=Number of kingsTotal number of cardsP(\text{King on 1st draw}) = \frac{\text{Number of kings}}{\text{Total number of cards}} P(King on 1st draw)=452P(\text{King on 1st draw}) = \frac{4}{52} P(King on 1st draw)=113P(\text{King on 1st draw}) = \frac{1}{13}

  3. Calculate the probability of drawing a king on the second draw.

    Since the first card is replaced, the deck returns to its original state (52 cards, 4 kings). Therefore, the probability of drawing a king on the second draw is the same as the first draw.

    P(King on 2nd draw)=Number of kingsTotal number of cardsP(\text{King on 2nd draw}) = \frac{\text{Number of kings}}{\text{Total number of cards}} P(King on 2nd draw)=452P(\text{King on 2nd draw}) = \frac{4}{52} P(King on 2nd draw)=113P(\text{King on 2nd draw}) = \frac{1}{13}

  4. Calculate the probability that both cards are kings.

    Since the draws are independent events (due to replacement), the probability of both events occurring is the product of their individual probabilities.

    P(Both Kings)=P(King on 1st draw)×P(King on 2nd draw)P(\text{Both Kings}) = P(\text{King on 1st draw}) \times P(\text{King on 2nd draw}) P(Both Kings)=113×113P(\text{Both Kings}) = \frac{1}{13} \times \frac{1}{13} P(Both Kings)=1169P(\text{Both Kings}) = \frac{1}{169}

    The probability that the two cards are both kings is 1169\frac{1}{169}.

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