Let `Aa n dB` are events of an experiment and `P(A)=1//4, P(AuuB)=1//2,` then value of `P(B//A^c)` is `2//3` b. `1//3` c. `5//6` d. `1//2`

Options

A. `2//3`
B. `1//3`
C. `5//6`
D. `1//2`

Explanation

To solve the problem, we need to find the value of \( P(B | A^c) \) using the given probabilities \( P(A) = \frac{1}{4} \) and \( P(A \cup B) = \frac{1}{2} \).

### Step-by-step Solution:

1. **Understand the Problem**: We need to find \( P(B | A^c) \), which is the probability of event B occurring given that event A does not occur.

2. **Use the Conditional Probability Formula**: The formula for conditional probability is:
\[ P(B | A^c) = \frac{P(B \cap A^c)}{P(A^c)} \]

3. **Calculate \( P(A^c) \)**: Since \( P(A) = \frac{1}{4} \), we can find \( P(A^c) \) as follows:
\[ P(A^c) = 1 - P(A) = 1 - \frac{1}{4} = \frac{3}{4} \]

4. **Find \( P(B \cap A^c) \)**: We can express \( P(B \cap A^c) \) using the relationship between \( P(A \cup B) \), \( P(A) \), and \( P(B) \):
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Rearranging gives:
\[ P(B) = P(A \cup B) - P(A) + P(A \cap B) \]

However, we do not have \( P(B) \) or \( P(A \cap B) \) directly. Instead, we can express \( P(B \cap A^c) \) as:
\[ P(B \cap A^c) = P(B) - P(A \cap B) \]
But we need \( P(A \cap B) \) to proceed.

5. **Express \( P(A \cup B) \)**: From the given data:
\[ P(A \cup B) = \frac{1}{2} \]
Thus, we can write:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting the known values:
\[ \frac{1}{2} = \frac{1}{4} + P(B) - P(A \cap B) \]

6. **Rearranging the Equation**: Rearranging gives:
\[ P(B) - P(A \cap B) = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \]
Therefore, we have:
\[ P(B \cap A^c) = P(B) - P(A \cap B) = \frac{1}{4} \]

7. **Substituting Back into the Conditional Probability Formula**:
\[ P(B | A^c) = \frac{P(B \cap A^c)}{P(A^c)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{4} \times \frac{4}{3} = \frac{1}{3} \]

8. **Final Answer**: Thus, the value of \( P(B | A^c) \) is:
\[ \boxed{\frac{1}{3}} \]