From a log of 6 items containing 2 defective items, a sample of 4 items are drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn without replacement, find :
(a) The probability distribution of X.
(b) Mean of X.
(c ) Variance of X.

Answer

To solve the problem step by step, we will find the probability distribution of the random variable \( X \), which denotes the number of defective items in a sample of 4 drawn from a log of 6 items containing 2 defective items. We will then calculate the mean and variance of \( X \).

### Step 1: Define the Random Variable \( X \)
The random variable \( X \) can take the values 0, 1, or 2, representing the number of defective items in the sample of 4 items drawn from the log.

### Step 2: Calculate the Total Combinations
The total number of ways to choose 4 items from 6 is given by the combination formula:
\[ \text{Total combinations} = \binom{6}{4} = 15 \]

### Step 3: Calculate the Probability for Each Value of \( X \)

#### (a) Probability Distribution of \( X \)

1. **Probability \( P(X = 0) \)**: This is the probability of selecting 0 defective items.
- We need to choose 0 defective items from the 2 defective items and 4 good items from the 4 good items.
\[ P(X = 0) = \frac{\binom{2}{0} \cdot \binom{4}{4}}{\binom{6}{4}} = \frac{1 \cdot 1}{15} = \frac{1}{15} \]

2. **Probability \( P(X = 1) \)**: This is the probability of selecting 1 defective item.
- We need to choose 1 defective item from the 2 defective items and 3 good items from the 4 good items.
\[ P(X = 1) = \frac{\binom{2}{1} \cdot \binom{4}{3}}{\binom{6}{4}} = \frac{2 \cdot 4}{15} = \frac{8}{15} \]

3. **Probability \( P(X = 2) \)**: This is the probability of selecting 2 defective items.
- We need to choose 2 defective items from the 2 defective items and 2 good items from the 4 good items.
\[ P(X = 2) = \frac{\binom{2}{2} \cdot \binom{4}{2}}{\binom{6}{4}} = \frac{1 \cdot 6}{15} = \frac{6}{15} = \frac{2}{5} \]

Now we can summarize the probability distribution of \( X \):
\[ \begin{array}{|c|c|} \hline X & P(X) \\ \hline 0 & \frac{1}{15} \\ 1 & \frac{8}{15} \\ 2 & \frac{6}{15} \\ \hline \end{array} \]

### Step 4: Calculate the Mean of \( X \)
The mean \( E(X) \) is calculated using the formula:
\[ E(X) = \sum_{i=0}^{2} x_i P(X = x_i) \]
Calculating each term:
\[ E(X) = 0 \cdot \frac{1}{15} + 1 \cdot \frac{8}{15} + 2 \cdot \frac{6}{15} = 0 + \frac{8}{15} + \frac{12}{15} = \frac{20}{15} = \frac{4}{3} \]

### Step 5: Calculate the Variance of \( X \)
The variance \( Var(X) \) is calculated using the formula:
\[ Var(X) = E(X^2) - (E(X))^2 \]
First, we calculate \( E(X^2) \):
\[ E(X^2) = \sum_{i=0}^{2} x_i^2 P(X = x_i) = 0^2 \cdot \frac{1}{15} + 1^2 \cdot \frac{8}{15} + 2^2 \cdot \frac{6}{15} \]
Calculating each term:
\[ E(X^2) = 0 + \frac{8}{15} + \frac{24}{15} = \frac{32}{15} \]
Now, substitute into the variance formula:
\[ Var(X) = E(X^2) - (E(X))^2 = \frac{32}{15} - \left(\frac{4}{3}\right)^2 = \frac{32}{15} - \frac{16}{9} \]
Finding a common denominator (which is 45):
\[ Var(X) = \frac{32 \cdot 3}{45} - \frac{16 \cdot 5}{45} = \frac{96 - 80}{45} = \frac{16}{45} \]

### Final Results
- **Probability Distribution**:
\[ P(X = 0) = \frac{1}{15}, \quad P(X = 1) = \frac{8}{15}, \quad P(X = 2) = \frac{2}{5} \]
- **Mean of \( X \)**: \( \frac{4}{3} \)
- **Variance of \( X \)**: \( \frac{16}{45} \)