Question

Two cards are drawn simultaneously from a well shuffled pack of 52 cards. If X is the random variable of getting queens, then the value of $2 E(X) + 3 E(X^2)$ for the number of queens is

• A. $\frac{132}{221}$
• B. $\frac{108}{221}$
• C. $\frac{176}{221}$
• D. $\frac{68}{221}$

Hint:

For drawing without replacement, $E(X) = n \cdot \frac{M}{N}$ where $n$ is draws, $M$ is successes in deck, and $N$ is total cards.

Answer 1

The Correct Option is A

Step 1: Concept
This is a hypergeometric distribution problem where $X$ can be 0, 1, or 2.

Step 2: Meaning
Calculate $P(X=0), P(X=1), P(X=2)$ using combinations $\binom{n}{r}$. Total outcomes $= \binom{52}{2} = \frac{52 \times 51}{2} = 1326$.

Step 3: Analysis
$P(X=1) = \frac{\binom{4}{1} \binom{48}{1}}{1326} = \frac{192}{1326}$. $P(X=2) = \frac{\binom{4}{2} \binom{48}{0}}{1326} = \frac{6}{1326}$. $E(X) = 0 \cdot P(0) + 1 \cdot \frac{192}{1326} + 2 \cdot \frac{6}{1326} = \frac{204}{1326} = \frac{2}{13}$. $E(X^2) = 0^2 \cdot P(0) + 1^2 \cdot \frac{192}{1326} + 2^2 \cdot \frac{6}{1326} = \frac{192 + 24}{1326} = \frac{216}{1326} = \frac{36}{221}$. Value $= 2(\frac{2}{13}) + 3(\frac{36}{221}) = \frac{4}{13} + \frac{108}{221} = \frac{68 + 108}{221}$.

Step 4: Conclusion
Value $= \frac{176}{221}$ ... wait, re-checking sum. $68+108 = 176$. Matches (C). Final Answer: (C)