Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let X denote the random variable of number of aces obtained in the two drawn cards. Then P(X = 1) + P(X = 2) equals

Options

A. `52//169`
B. `25//169`
C. `49//169`
D. `24//169`

Answer

To solve the problem, we need to calculate \( P(X = 1) + P(X = 2) \), where \( X \) is the random variable representing the number of aces obtained when drawing two cards with replacement from a standard deck of 52 cards.

### Step-by-Step Solution:

1. **Identify the Total Number of Aces and Non-Aces:**
- In a standard deck of 52 cards, there are 4 aces and 48 non-aces.

2. **Calculate the Probability of Drawing an Ace:**
- The probability of drawing an ace (denote as \( p \)) is:
\[ p = \frac{4}{52} = \frac{1}{13} \]

3. **Calculate the Probability of Drawing a Non-Ace:**
- The probability of drawing a non-ace (denote as \( q \)) is:
\[ q = 1 - p = 1 - \frac{1}{13} = \frac{12}{13} \]

4. **Calculate \( P(X = 1) \):**
- \( P(X = 1) \) is the probability of getting exactly 1 ace in 2 draws. This can happen in two ways: Ace on the first draw and non-ace on the second, or non-ace on the first draw and ace on the second.
- The formula for this is:
\[ P(X = 1) = \binom{2}{1} p^1 q^1 = 2 \cdot \left(\frac{1}{13}\right)^1 \cdot \left(\frac{12}{13}\right)^1 \]
- Calculating this gives:
\[ P(X = 1) = 2 \cdot \frac{1}{13} \cdot \frac{12}{13} = \frac{24}{169} \]

5. **Calculate \( P(X = 2) \):**
- \( P(X = 2) \) is the probability of getting 2 aces in 2 draws. This can only happen if both draws result in an ace:
- The formula for this is:
\[ P(X = 2) = \binom{2}{2} p^2 q^0 = 1 \cdot \left(\frac{1}{13}\right)^2 \cdot 1 \]
- Calculating this gives:
\[ P(X = 2) = \left(\frac{1}{13}\right)^2 = \frac{1}{169} \]

6. **Combine the Probabilities:**
- Now, we can find \( P(X = 1) + P(X = 2) \):
\[ P(X = 1) + P(X = 2) = \frac{24}{169} + \frac{1}{169} = \frac{25}{169} \]

### Final Answer:
\[ P(X = 1) + P(X = 2) = \frac{25}{169} \]