If A and B are two events such that P(A) `gt` 0 and P(B) `lt` 1, then P(`A|barB`) is equal to

Options

A. `1-P(barA|B)`
B. `1-P(A|B)`
C. `(P(barA))/(P(B))`
D. `1-P(barA|barB)`

Answer

To solve the problem of finding \( P(A | \bar{B}) \) given that \( P(A) > 0 \) and \( P(B) < 1 \), we can use the definition of conditional probability and some properties of probabilities.

### Step-by-Step Solution:

1. **Understanding Conditional Probability**:
The conditional probability \( P(A | \bar{B}) \) is defined as:
\[ P(A | \bar{B}) = \frac{P(A \cap \bar{B})}{P(\bar{B})} \]

2. **Finding \( P(\bar{B}) \)**:
Since \( P(B) < 1 \), we can find \( P(\bar{B}) \) as:
\[ P(\bar{B}) = 1 - P(B) \]

3. **Using the Law of Total Probability**:
We can express \( P(A) \) in terms of \( P(A \cap B) \) and \( P(A \cap \bar{B}) \):
\[ P(A) = P(A \cap B) + P(A \cap \bar{B}) \]

4. **Rearranging for \( P(A \cap \bar{B}) \)**:
From the equation above, we can rearrange to find \( P(A \cap \bar{B}) \):
\[ P(A \cap \bar{B}) = P(A) - P(A \cap B) \]

5. **Substituting into the Conditional Probability Formula**:
Now we substitute \( P(A \cap \bar{B}) \) back into the conditional probability formula:
\[ P(A | \bar{B}) = \frac{P(A) - P(A \cap B)}{P(\bar{B})} \]

6. **Expressing \( P(A \cap B) \)**:
We can express \( P(A \cap B) \) in terms of \( P(A) \) and \( P(B) \) if we have additional information about independence or joint probabilities, but for now, we keep it as is.

7. **Final Expression**:
The final expression for \( P(A | \bar{B}) \) is:
\[ P(A | \bar{B}) = \frac{P(A) - P(A \cap B)}{1 - P(B)} \]

### Conclusion:
Thus, we have derived the expression for \( P(A | \bar{B}) \) in terms of \( P(A) \) and \( P(B) \).