If A and B are two events that `P(A) gt 0 and P(B) != 1 ` , then P(A/B) is equal to

Options

A. `1-P(A//B)`
B. `1-P(A//B)`
C. `(1-P(A cup B))/(P(B))`
D. `(P(A))/(P(B))`

Answer

To solve the problem, we need to find the conditional probability \( P(A|B) \), which is the probability of event A occurring given that event B has occurred.

### Step-by-Step Solution:

1. **Understanding Conditional Probability**:
The conditional probability \( P(A|B) \) is defined as:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
where \( P(A \cap B) \) is the probability that both events A and B occur, and \( P(B) \) is the probability that event B occurs.

2. **Given Conditions**:
We are given that \( P(A) > 0 \) and \( P(B) \neq 1 \). This means that event A has a positive probability of occurring, and event B does not always occur.

3. **Using the Definition**:
From the definition of conditional probability, we can express \( P(A|B) \) as:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]

4. **Finding \( P(A \cap B) \)**:
We can also express \( P(A \cap B) \) in terms of the union of A and B. We know that:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Rearranging this gives us:
\[ P(A \cap B) = P(A) + P(B) - P(A \cup B) \]

5. **Substituting Back**:
Substituting \( P(A \cap B) \) back into the conditional probability formula gives us:
\[ P(A|B) = \frac{P(A) + P(B) - P(A \cup B)}{P(B)} \]

6. **Final Expression**:
Thus, we can express \( P(A|B) \) as:
\[ P(A|B) = \frac{P(A) + P(B) - P(A \cup B)}{P(B)} \]

### Conclusion:
So, the final answer for \( P(A|B) \) is:
\[ P(A|B) = \frac{P(A) + P(B) - P(A \cup B)}{P(B)} \]