Multiple choice

Four cards are drawn at random from a well-shuffled pack of 52 cards. Find the probability that there is exactly one pair.

  1. $\dfrac{(^{13}C_1 \times ^4C_2 \times ^{48}C_1 \times ^{44}C_1)}{^{52}C_4}$
  2. $\dfrac{(^{13}C_1 \times ^4C_2 \times ^{48}C_1 )}{^{52}C_4}$
  3. $\dfrac{(^{13}C_1 \times ^4C_2 \times ^{48}C_1 \times ^{44}C_1)}{^{48}C_4}$
  4. $\dfrac{(^{13}C_1 \times ^{48}C_1 \times ^{44}C_1)}{^{52}C_4}$
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A Correct answer
AI explanation

To find the probability of exactly one pair in a four-card hand, the total number of outcomes is choosing 4 cards from 52, written as 52C4. For the favorable outcomes, select the rank for the pair in 13C1 ways, choose 2 suits for that rank in 4C2 ways, and then pick the remaining two cards of different ranks in 48C1 and 44C1 ways respectively. The probability is the ratio of the favorable combinations to the total combinations, yielding the formula (13C1 multiplied by 4C2 multiplied by 48C1 multiplied by 44C1) divided by 52C4. The result is (13C1 multiplied by 4C2 multiplied by 48C1 multiplied by 44C1) divided by 52C4.

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