If A and B are independent events such that `P(A) gt 0, P(B) gt 0`, then

Options

A. A and B are mutually exclusive
B. A and `overline(B)` are independent
C. `A (A cup B)=P(overline(A))P(overline(B))`
D. `P(A//B)=P(overline(A)//B)`

Answer

To solve the problem, we need to show that if A and B are independent events, then A and B bar (the complement of B) are also independent.

### Step-by-step Solution:

1. **Understand the Definitions**:
- Independent events A and B mean that:
\[ P(A \cap B) = P(A) \cdot P(B) \]
- The complement of event B, denoted as \( B' \) or \( B^c \), is the event that B does not occur.

2. **Find \( P(A \cap B') \)**:
- We can express \( P(A \cap B') \) using the formula for the intersection of A and the complement of B:
\[ P(A \cap B') = P(A) - P(A \cap B) \]
- Since A and B are independent, we substitute \( P(A \cap B) \):
\[ P(A \cap B') = P(A) - P(A) \cdot P(B) \]

3. **Factor Out \( P(A) \)**:
- From the equation above, we can factor out \( P(A) \):
\[ P(A \cap B') = P(A) \cdot (1 - P(B)) \]
- Recognizing that \( 1 - P(B) = P(B') \), we rewrite the equation:
\[ P(A \cap B') = P(A) \cdot P(B') \]

4. **Conclusion**:
- The equation \( P(A \cap B') = P(A) \cdot P(B') \) shows that A and B' are independent events.
- Therefore, we conclude that if A and B are independent, then A and B' are also independent.

### Final Answer:
The correct conclusion is that A and B' are independent events.