Question

Two cards are drawn successively with replacement from fair playing 52 cards. let X denote number of kings obtained when two cards are drawn, then ( E(X^2) = )

• A. (\frac{24}{169})
• B. (\frac{26}{169})
• C. (\frac{27}{169})
• D. [suspicious link removed]

Hint:

For replacement, probabilities remain constant for each draw.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
This is a binomial distribution problem since cards are replaced. We need to find the expected value of the square of the random variable \( X \).
Step 3: Detailed Explanation:
Number of trials \( n = 2 \).
Probability of a king \( p = \frac{4}{52} = \frac{1}{13} \).
Probability of not a king \( q = \frac{12}{13} \).
The distribution of \( X \) is:
- \( P(X=0) = q^2 = \left(\frac{12}{13}\right)^2 = \frac{144}{169} \)
- \( P(X=1) = 2pq = 2\left(\frac{1}{13}\right)\left(\frac{12}{13}\right) = \frac{24}{169} \)
- \( P(X=2) = p^2 = \left(\frac{1}{13}\right)^2 = \frac{1}{169} \)
Calculation of \( E(X^2) \):
\[ E(X^2) = \sum x_i^2 P(x_i) \]
\[ E(X^2) = 0^2\left(\frac{144}{169}\right) + 1^2\left(\frac{24}{169}\right) + 2^2\left(\frac{1}{169}\right) \]
\[ E(X^2) = 0 + \frac{24}{169} + \frac{4}{169} = \frac{28}{169} \]
Step 4: Final Answer:
The value of \( E(X^2) \) is \( \frac{28}{169} \).