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: Understanding the Concept:
We are drawing 2 cards without replacement from a deck of 52. The random variable $X$ denotes the number of queens drawn.
$X$ can take values 0, 1, or 2.
We need to find the probability distribution of $X$ to calculate the expected values $E(X)$ and $E(X^2)$.
Step 2: Key Formula or Approach:
The probability $P(X=x)$ is calculated using combinations: $\frac{\text{favorable outcomes}}{\text{total outcomes}}$.
Total number of queens = 4, non-queens = 48.
$E(X) = \sum x \cdot P(X=x)$
$E(X^2) = \sum x^2 \cdot P(X=x)$
Step 3: Detailed Explanation:
Let's find the probabilities for each value of $X$.
Total ways to draw 2 cards from 52 is $\binom{52}{2} = \frac{52 \times 51}{2 \times 1} = 26 \times 51 = 1326$.
- $P(X=0)$: Drawing 0 queens means drawing 2 non-queens from 48.
$P(X=0) = \frac{\binom{48}{2}}{\binom{52}{2}} = \frac{\frac{48 \times 47}{2}}{1326} = \frac{24 \times 47}{1326} = \frac{1128}{1326} = \frac{188}{221}$ (dividing by 6)
- $P(X=1)$: Drawing 1 queen from 4 and 1 non-queen from 48.
$P(X=1) = \frac{\binom{4}{1} \times \binom{48}{1}}{\binom{52}{2}} = \frac{4 \times 48}{1326} = \frac{192}{1326} = \frac{32}{221}$ (dividing by 6)
- $P(X=2)$: Drawing 2 queens from 4.
$P(X=2) = \frac{\binom{4}{2}}{\binom{52}{2}} = \frac{\frac{4 \times 3}{2}}{1326} = \frac{6}{1326} = \frac{1}{221}$ (dividing by 6)
Let's check if probabilities sum to 1: $\frac{188 + 32 + 1}{221} = \frac{221}{221} = 1$. Correct.
Now calculate $E(X)$ and $E(X^2)$:
\[ E(X) = \sum_{x=0}^{2} x \cdot P(x) = 0\left(\frac{188}{221}\right) + 1\left(\frac{32}{221}\right) + 2\left(\frac{1}{221}\right) = \frac{32 + 2}{221} = \frac{34}{221} \]
\[ E(X^2) = \sum_{x=0}^{2} x^2 \cdot P(x) = 0^2\left(\frac{188}{221}\right) + 1^2\left(\frac{32}{221}\right) + 2^2\left(\frac{1}{221}\right) = \frac{32 + 4}{221} = \frac{36}{221} \]
Finally, evaluate the required expression:
\[ 2E(X) + 3E(X^2) = 2\left(\frac{34}{221}\right) + 3\left(\frac{36}{221}\right) \]
\[ = \frac{68}{221} + \frac{108}{221} = \frac{176}{221} \]
Step 4: Final Answer:
The value is $\frac{176}{221}$.