A lot contains equal number of defective and non defective bulbs. Two bulbs ar drawn at random, one at a time, with replacement. The events A, B, C are defined as
A : The first bulb is defective
B : The second bulb is non defective
C : Two bulbs are either both defective and non defective.
Statement-1 : A, B, C are pairwise independent.
Statement-2 : A, B, C are mutually independent.

Options

A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for statement -1
B. Statement-1 is true, Statement-2 is true, Statement-2 is Not a correct explanation for statement -1
C. Statement-1 is true, Statement-2 is False
D. Statement-1 is False, Statement-2 is true

Answer

To solve the problem, we need to analyze the events A, B, and C defined in the question and determine their independence.

### Step-by-Step Solution:

1. **Define the Events:**
- Let \( A \): The event that the first bulb drawn is defective.
- Let \( B \): The event that the second bulb drawn is non-defective.
- Let \( C \): The event that both bulbs are either both defective or both non-defective.

2. **Determine Probabilities:**
- Since there are equal numbers of defective and non-defective bulbs, the probability of drawing a defective bulb (event \( A \)) is:
\[ P(A) = \frac{1}{2} \]
- Similarly, the probability of drawing a non-defective bulb (event \( B \)) is:
\[ P(B) = \frac{1}{2} \]
- For event \( C \), we need to calculate the probability of both bulbs being either both defective or both non-defective:
- The probability of both bulbs being defective:
\[ P(A \cap A) = P(A) \times P(A) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]
- The probability of both bulbs being non-defective:
\[ P(B \cap B) = P(B) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]
- Therefore, the probability of event \( C \) is:
\[ P(C) = P(A \cap A) + P(B \cap B) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \]

3. **Check Pairwise Independence:**
- For events to be pairwise independent, the following must hold true:
- \( P(A \cap B) = P(A) \times P(B) \)
- \( P(A \cap C) = P(A) \times P(C) \)
- \( P(B \cap C) = P(B) \times P(C) \)

- Calculate \( P(A \cap B) \):
\[ P(A \cap B) = P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

- Calculate \( P(A \cap C) \):
- \( C \) includes both defective and non-defective scenarios. Hence, \( P(A \cap C) \) is the probability of the first bulb being defective and both being the same:
\[ P(A \cap C) = P(A \cap A) = \frac{1}{4} \]
- Check:
\[ P(A) \times P(C) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

- Calculate \( P(B \cap C) \):
- Similarly, \( P(B \cap C) \) is the probability of the first bulb being non-defective and both being the same:
\[ P(B \cap C) = P(B \cap B) = \frac{1}{4} \]
- Check:
\[ P(B) \times P(C) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

- Since all pairwise independence conditions are satisfied, events \( A, B, C \) are pairwise independent.

4. **Check Mutual Independence:**
- For mutual independence, we need:
\[ P(A \cap B \cap C) = P(A) \times P(B) \times P(C) \]
- However, \( P(A \cap B \cap C) \) cannot be calculated directly as it requires specific combinations of the events. Since we have established that \( A \) and \( B \) can occur independently of \( C \) but not necessarily together, we find that they are not mutually independent.

### Conclusion:
- **Statement 1**: True (A, B, C are pairwise independent).
- **Statement 2**: False (A, B, C are not mutually independent).