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

Options

A. (a)`1-P(A//B)`
B. (b)`1-P(barA//B)`
C. (c)`(1-P(A cup B))/(P(barB))`
D. (d)`(P(A))/(P(barB))`

Explanation

To solve the problem, we need to find \( P(\bar{A} | \bar{B}) \), where \( \bar{A} \) and \( \bar{B} \) are the complements of events \( A \) and \( B \), respectively.

### Step-by-Step Solution:

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

2. **Express \( P(\bar{A} \cap \bar{B}) \)**:
We can use the relationship between complements and unions:
\[ P(\bar{A} \cap \bar{B}) = 1 - P(A \cup B) \]
This is because the probability of the complement of the union of two events is equal to 1 minus the probability of the union.

3. **Substituting into the Conditional Probability Formula**:
Now we can substitute this expression back into our conditional probability formula:
\[ P(\bar{A} | \bar{B}) = \frac{1 - P(A \cup B)}{P(\bar{B})} \]

4. **Express \( P(\bar{B}) \)**:
Since \( P(B) \neq 1 \), we can express \( P(\bar{B}) \) as:
\[ P(\bar{B}) = 1 - P(B) \]

5. **Final Expression**:
Thus, we can rewrite our conditional probability as:
\[ P(\bar{A} | \bar{B}) = \frac{1 - P(A \cup B)}{1 - P(B)} \]

### Conclusion:
The final result for \( P(\bar{A} | \bar{B}) \) is:
\[ P(\bar{A} | \bar{B}) = \frac{1 - P(A \cup B)}{1 - P(B)} \]