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

Explanation

To find \( P(\bar{A} | \bar{B}) \), we can use the definition of conditional probability:

\[ P(\bar{A} | \bar{B}) = \frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})} \]

### Step 1: Identify the components
We know that:
- \( P(A) > 0 \)
- \( P(B) \neq 1 \)

### Step 2: Express \( P(\bar{A} \cap \bar{B}) \)
Using the property of complements, we can express \( P(\bar{A} \cap \bar{B}) \) in terms of unions:
\[ P(\bar{A} \cap \bar{B}) = P(\overline{A \cup B}) \]
This is because the complement of the union of two events is the intersection of their complements.

### Step 3: Use the complement rule
Using the complement rule, we have:
\[ P(\overline{A \cup B}) = 1 - P(A \cup B) \]

### Step 4: Substitute back into the conditional probability formula
Now, substituting this back into our expression for conditional probability, we get:
\[ P(\bar{A} | \bar{B}) = \frac{1 - P(A \cup B)}{P(\bar{B})} \]

### Step 5: Express \( P(\bar{B}) \)
Since \( P(B) \neq 1 \), we can express \( P(\bar{B}) \) as:
\[ P(\bar{B}) = 1 - P(B) \]

### Final Expression
Thus, we can write:
\[ P(\bar{A} | \bar{B}) = \frac{1 - P(A \cup B)}{1 - P(B)} \]

### Conclusion
This is the final expression for \( P(\bar{A} | \bar{B}) \).

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