A sample of 2 itmes is selected at random from a bag containing 5 items of which 2 are defective. Then mean of number of defective items is

Options

A. `(4)/(5)`
B. `(1)/(5)`
C. `(2)/(5)`
D. `(3)/(5)`

Answer

To find the mean of the number of defective items when a sample of 2 items is selected at random from a bag containing 5 items (2 of which are defective), we can follow these steps:

### Step 1: Define the Random Variable
Let \( X \) be the random variable representing the number of defective items in the sample of 2 items selected from the bag. The possible values of \( X \) are 0, 1, or 2.

### Step 2: Calculate the Total Number of Ways to Select 2 Items
The total number of ways to select 2 items from 5 items is given by the combination formula:
\[ C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

### Step 3: Calculate the Probability of Each Case
1. **Case 1: No defective items (X = 0)**
- Number of ways to choose 2 non-defective items from 3:
\[ C(3, 2) = \frac{3!}{2!(3-2)!} = \frac{3 \times 2}{2 \times 1} = 3 \]
- Probability:
\[ P(X = 0) = \frac{C(3, 2)}{C(5, 2)} = \frac{3}{10} \]

2. **Case 2: One defective item (X = 1)**
- Number of ways to choose 1 defective item from 2 and 1 non-defective item from 3:
\[ C(2, 1) \times C(3, 1) = 2 \times 3 = 6 \]
- Probability:
\[ P(X = 1) = \frac{6}{10} = \frac{3}{5} \]

3. **Case 3: Two defective items (X = 2)**
- Number of ways to choose 2 defective items from 2:
\[ C(2, 2) = 1 \]
- Probability:
\[ P(X = 2) = \frac{1}{10} \]

### Step 4: Calculate the Mean of the Random Variable
The mean (expected value) \( E(X) \) is calculated using the formula:
\[ E(X) = \sum (X_i \cdot P(X_i)) \]
Substituting the values we found:
\[ E(X) = 0 \cdot P(X = 0) + 1 \cdot P(X = 1) + 2 \cdot P(X = 2) \]
\[ E(X) = 0 \cdot \frac{3}{10} + 1 \cdot \frac{3}{5} + 2 \cdot \frac{1}{10} \]
\[ E(X) = 0 + \frac{3}{5} + \frac{2}{10} \]
\[ E(X) = \frac{3}{5} + \frac{1}{5} = \frac{4}{5} \]

### Final Answer
The mean of the number of defective items is \( \frac{4}{5} \).

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