Question

Let two cards be drawn at random from a pack of 52 playing cards. Let X be the number of aces obtained. Then the value of E(X) is

• A. $\frac{5}{13}$
• B. $\frac{1}{13}$
• C. $\frac{2}{13}$
• D. $\frac{37}{221}$

Hint:

Linearity of expectation is an incredible mathematical superpower! It holds completely true whether the events are independent or dependent (like drawing without replacement). For drawing $n$ cards, the expected number of aces is always simply $n \times \frac{4}{52}$. For 2 cards, it's $2 \times \frac{1}{13} = \frac{2}{13}$ in a single step!

Answer 1

The Correct Option is C

Step 1: Define the random variable clearly.
$X$ counts the aces among the two drawn cards, so $X$ can be $0$, $1$, or $2$. We want $E(X) = \sum x\,P(X=x)$.
Step 2: Find $P(X=0)$.
Choose 2 non-ace cards from 48: $P(X=0) = \dfrac{\binom{48}{2}}{\binom{52}{2}} = \dfrac{1128}{1326}$.
Step 3: Find $P(X=1)$.
One ace from 4 and one non-ace from 48: $P(X=1) = \dfrac{\binom{4}{1}\binom{48}{1}}{\binom{52}{2}} = \dfrac{192}{1326}$.
Step 4: Find $P(X=2)$.
Both aces from 4: $P(X=2) = \dfrac{\binom{4}{2}}{\binom{52}{2}} = \dfrac{6}{1326}$.
Step 5: Plug into the expectation sum.
$E(X) = 0\cdot\dfrac{1128}{1326} + 1\cdot\dfrac{192}{1326} + 2\cdot\dfrac{6}{1326} = \dfrac{192 + 12}{1326} = \dfrac{204}{1326}$.
Step 6: Simplify the fraction.
$\dfrac{204}{1326} = \dfrac{2}{13}$ after dividing top and bottom by $102$.
\[ \boxed{E(X) = \dfrac{2}{13}} \]