Question

From a well shuffled pack of 52 cards, two cards are drawn at random. Then, the probability of both the cards being kings is

• A. \( \frac{1}{15} \)
• B. \( \frac{25}{57} \)
• C. \( \frac{35}{256} \)
• D. \( \frac{1}{221} \)

Hint:

Probability = (Number of favorable outcomes) / (Total number of possible outcomes).
Use combinations \(\binom{n}{k}\) when selecting items without regard to order.
Alternatively, use sequential probability: P(A and B) = P(A) * P(B|A).

Answer 1

The Correct Option is D

Total number of cards in a pack = 52. Number of kings in a pack = 4. We are drawing two cards at random. The total number of ways to choose 2 cards from 52 is \(\binom{52}{2}\). \[ \binom{52}{2} = \frac{52 \times 51}{2 \times 1} = 26 \times 51 = 1326 \] The number of ways to choose 2 kings from the 4 kings is \(\binom{4}{2}\). \[ \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 \] The probability of both cards being kings is: \[ P(\text{both kings}) = \frac{\text{Number of ways to choose 2 kings}}{\text{Total number of ways to choose 2 cards}} = \frac{\binom{4}{2}}{\binom{52}{2}} \] \[ P(\text{both kings}) = \frac{6}{1326} \] Simplify the fraction: Divide by 6: \(6 \div 6 = 1\). \(1326 \div 6\): \(13 \div 6 = 2\) remainder 1. \(12 \div 6 = 2\) remainder 0. \(6 \div 6 = 1\) remainder 0. So, \(1326 \div 6 = 221\). \[ P(\text{both kings}) = \frac{1}{221} \] This matches option (d). Alternatively, using conditional probability: P(1st card is king) = \(4/52\). Given 1st was a king, P(2nd card is king) = \(3/51\). P(both kings) = \(P(\text{1st is king}) \times P(\text{2nd is king | 1st is king})\) \( = \frac{4}{52} \times \frac{3}{51} = \frac{1}{13} \times \frac{1}{17} = \frac{1}{13 \times 17} = \frac{1}{221} \). \[ \boxed{\frac{1}{221}} \]