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: Understanding the Question:
Two cards are sampled simultaneously from a standard deck. We are tracking a discrete random variable $X$ representing the total count of aces drawn, and we need to calculate its mathematical expectation E(X).

Step 2: Key Formula or Approach:
The expected value of a discrete random variable is computed as $\text{E(X)} = \sum x_i \cdot P(x_i)$. Alternatively, we can use the linear properties of expectation. Sampling cards can be viewed as a sequence of dependent events, or we can use indicators where the expectation of a sum of random variables equals the sum of their individual expectations.

Step 3: Detailed Explanation:
Let's use a highly elegant probability method based on the linearity of expectation:
Let $X_1$ be an indicator random variable for the first card drawn ($X_1 = 1$ if it's an ace, $0$ otherwise).
Let $X_2$ be an indicator random variable for the second card drawn ($X_2 = 1$ if it's an ace, $0$ otherwise).
The total number of aces is $X = X_1 + X_2$. By linearity of expectation:
$$\text{E(X)} = \text{E}(X_1) + \text{E}(X_2)$$ The probability that any single card selected at random from a fresh deck is an ace is:
$$\text{P}(\text{Ace}) = \frac{4}{52} = \frac{1}{13}$$ Therefore, the expected value for each individual card indicator is:
$$\text{E}(X_1) = \frac{1}{13} \quad \text{and} \quad \text{E}(X_2) = \frac{1}{13}$$ Summing these individual expectations together gives:
$$\text{E(X)} = \frac{1}{13} + \frac{1}{13} = \frac{2}{13}$$ This results in exactly $\frac{2}{13}$, which matches option (C).

Step 4: Final Answer:
The expected value E(X) is $\frac{2}{13}$, which corresponds to option (C).