A box contains 6 red balls and 2 black balls. Two balls are drawn at random, from it without replacement. If X denotes the number of red balls drawn then E(X) is equal to:

Options

A. `3//2`
B. `1//2`
C. `5//2`
D. `27//28`

Answer

To find the expected value \( E(X) \) of the random variable \( X \), which denotes the number of red balls drawn when two balls are drawn from a box containing 6 red balls and 2 black balls, we can follow these steps:

### Step 1: Identify Possible Values of \( X \)
The random variable \( X \) can take the following values:
- \( X = 0 \): No red balls are drawn (both balls are black).
- \( X = 1 \): One red ball and one black ball are drawn.
- \( X = 2 \): Two red balls are drawn.

### Step 2: Calculate Probabilities for Each Value of \( X \)

#### Probability of \( X = 0 \)
To find the probability of drawing 0 red balls (i.e., drawing 2 black balls), we can use the combination formula:
\[ P(X = 0) = \frac{\text{Number of ways to choose 2 black balls}}{\text{Total ways to choose 2 balls from 8}} \]
Calculating this:
\[ P(X = 0) = \frac{\binom{2}{2}}{\binom{8}{2}} = \frac{1}{\frac{8 \times 7}{2}} = \frac{1}{28} \]

#### Probability of \( X = 1 \)
For the case of drawing 1 red ball and 1 black ball:
\[ P(X = 1) = \frac{\text{Number of ways to choose 1 red and 1 black}}{\text{Total ways to choose 2 balls from 8}} \]
Calculating this:
\[ P(X = 1) = \frac{\binom{6}{1} \cdot \binom{2}{1}}{\binom{8}{2}} = \frac{6 \cdot 2}{28} = \frac{12}{28} = \frac{3}{7} \]

#### Probability of \( X = 2 \)
For the case of drawing 2 red balls:
\[ P(X = 2) = \frac{\text{Number of ways to choose 2 red balls}}{\text{Total ways to choose 2 balls from 8}} \]
Calculating this:
\[ P(X = 2) = \frac{\binom{6}{2}}{\binom{8}{2}} = \frac{15}{28} \]

### Step 3: Calculate the Expected Value \( E(X) \)
The expected value is calculated using the formula:
\[ E(X) = \sum (x_i \cdot P(X = x_i)) \]
Substituting the values we calculated:
\[ E(X) = 0 \cdot P(X = 0) + 1 \cdot P(X = 1) + 2 \cdot P(X = 2) \]
\[ E(X) = 0 \cdot \frac{1}{28} + 1 \cdot \frac{12}{28} + 2 \cdot \frac{15}{28} \]
\[ E(X) = 0 + \frac{12}{28} + \frac{30}{28} = \frac{42}{28} = \frac{3}{2} \]

### Final Answer
Thus, the expected value \( E(X) \) is:
\[ E(X) = \frac{3}{2} \]