Two cards are drawn from 52 playing cards without replacement. The variance of the number of ace is

Options

A. `400/2873`
B. `600/2873`
C. `200/2873`
D. None of these

Explanation

To find the variance of the number of aces when drawing two cards from a standard deck of 52 playing cards without replacement, we can follow these steps:

### Step 1: Define the Random Variable
Let \( X \) be the random variable representing the number of aces drawn. The possible values for \( X \) are 0, 1, or 2.

### Step 2: Calculate the Probabilities
We need to calculate the probabilities for each value of \( X \).

#### Probability \( P(X = 0) \)
This is the probability of drawing 0 aces. We can choose 2 cards from the 48 non-ace cards.

\[ P(X = 0) = \frac{\binom{48}{2}}{\binom{52}{2}} \]

Calculating \( \binom{48}{2} \):

\[ \binom{48}{2} = \frac{48 \times 47}{2} = 1128 \]

Calculating \( \binom{52}{2} \):

\[ \binom{52}{2} = \frac{52 \times 51}{2} = 1326 \]

Thus,

\[ P(X = 0) = \frac{1128}{1326} = \frac{188}{221} \]

#### Probability \( P(X = 1) \)
This is the probability of drawing 1 ace. We can choose 1 ace from the 4 aces and 1 non-ace from the 48 non-aces.

\[ P(X = 1) = \frac{\binom{4}{1} \times \binom{48}{1}}{\binom{52}{2}} = \frac{4 \times 48}{1326} = \frac{192}{1326} = \frac{32}{221} \]

#### Probability \( P(X = 2) \)
This is the probability of drawing 2 aces. We can choose 2 aces from the 4 aces.

\[ P(X = 2) = \frac{\binom{4}{2}}{\binom{52}{2}} = \frac{6}{1326} = \frac{1}{221} \]

### Step 3: Summary of Probabilities
Now we summarize the probabilities:

- \( P(X = 0) = \frac{188}{221} \)
- \( P(X = 1) = \frac{32}{221} \)
- \( P(X = 2) = \frac{1}{221} \)

### Step 4: Calculate \( E(X) \)
The expected value \( E(X) \) is calculated as follows:

\[ E(X) = \sum x \cdot P(X = x) = 0 \cdot P(X = 0) + 1 \cdot P(X = 1) + 2 \cdot P(X = 2) \]

Calculating:

\[ E(X) = 0 + 1 \cdot \frac{32}{221} + 2 \cdot \frac{1}{221} = \frac{32}{221} + \frac{2}{221} = \frac{34}{221} \]

### Step 5: Calculate \( E(X^2) \)
Now we calculate \( E(X^2) \):

\[ E(X^2) = \sum x^2 \cdot P(X = x) = 0^2 \cdot P(X = 0) + 1^2 \cdot P(X = 1) + 2^2 \cdot P(X = 2) \]

Calculating:

\[ E(X^2) = 0 + 1 \cdot \frac{32}{221} + 4 \cdot \frac{1}{221} = \frac{32}{221} + \frac{4}{221} = \frac{36}{221} \]

### Step 6: Calculate Variance
The variance \( Var(X) \) is given by the formula:

\[ Var(X) = E(X^2) - (E(X))^2 \]

Substituting the values:

\[ Var(X) = \frac{36}{221} - \left(\frac{34}{221}\right)^2 \]

Calculating \( (E(X))^2 \):

\[ (E(X))^2 = \frac{34^2}{221^2} = \frac{1156}{48841} \]

Now substituting back into the variance formula:

\[ Var(X) = \frac{36 \cdot 221 - 1156}{221^2} = \frac{7956 - 1156}{48841} = \frac{6800}{48841} \]

### Final Answer
Thus, the variance of the number of aces drawn is:

\[ Var(X) = \frac{6800}{48841} \]