A lot contains 50 defective and 50 non-defectivebulbs. Two bulbs are drawn at random one at a time withreplacementevents A, and as first bul. The B C are defined theis defective, the second bulb is non-defective, the two banboth defective or non-defective, respectively. Then,(a) A, B and C are pairwise independent(b) A, B and C are pairwise not independent(c) A, B and C are independent(d) None of the above

Options

A. A and B are independent
B. B and C are independent
C. A and C are independent
D. A,B and C are pairwise independent

Explanation

To solve the problem, we need to analyze the events A, B, and C defined in the context of drawing bulbs from a lot containing 50 defective and 50 non-defective bulbs. The events are defined as follows:

- Event A: The first bulb drawn is defective.
- Event B: The second bulb drawn is non-defective.
- Event C: Both bulbs drawn are either defective or non-defective.

### Step 1: Calculate the probabilities of events A, B, and C

1. **Probability of Event A (First bulb is defective)**:
- There are 50 defective bulbs out of 100 total bulbs.
- \( P(A) = \frac{50}{100} = \frac{1}{2} \)

2. **Probability of Event B (Second bulb is non-defective)**:
- Again, there are 50 non-defective bulbs out of 100 total bulbs.
- \( P(B) = \frac{50}{100} = \frac{1}{2} \)

3. **Probability of Event C (Both bulbs are either defective or non-defective)**:
- The possible outcomes for C are:
- Both bulbs defective (X, X)
- Both bulbs non-defective (Y, Y)
- Calculate the probabilities:
- \( P(X, X) = P(A) \times P(A) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)
- \( P(Y, Y) = P(B) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)
- Therefore,
\[ P(C) = P(X, X) + P(Y, Y) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \]

### Step 2: Calculate joint probabilities for independence

1. **Joint Probability of A and B**:
- \( P(A \cap B) = P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)

2. **Joint Probability of A and C**:
- \( P(A \cap C) = P(X, X) + P(X, Y) = P(X, X) + 0 = \frac{1}{4} \)

3. **Joint Probability of B and C**:
- \( P(B \cap C) = P(Y, Y) + P(X, Y) = 0 + \frac{1}{4} = \frac{1}{4} \)

4. **Joint Probability of A, B, and C**:
- Since A and B are independent,
- \( P(A \cap B \cap C) = P(A \cap B) \times P(C) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \)

### Step 3: Check for independence

To check if A, B, and C are independent, we need to see if:
- \( P(A \cap B) = P(A) \times P(B) \)
- \( P(A \cap C) = P(A) \times P(C) \)
- \( P(B \cap C) = P(B) \times P(C) \)

From our calculations:
- \( P(A \cap B) = \frac{1}{4} \) and \( P(A) \times P(B) = \frac{1}{4} \) (independent)
- \( P(A \cap C) = \frac{1}{4} \) and \( P(A) \times P(C) = \frac{1}{4} \) (independent)
- \( P(B \cap C) = \frac{1}{4} \) and \( P(B) \times P(C) = \frac{1}{4} \) (independent)

Since all pairs are independent, we conclude that:
- A, B, and C are pairwise independent.

### Conclusion

Thus, the correct answer is:
**(a) A, B, and C are pairwise independent.**