A and B are two events such that `P(A cup B)=(3)/(4), P(A)=(1)/(3), P(overline(A) cap B)=`

Options

A. `(5)/(12)`
B. `(3)/(8)`
C. `(5)/(8)`
D. `(1)/(4)`

Explanation

To solve the problem, we need to find the probability of the event \( P(\overline{A} \cap B) \), given the following:

- \( P(A \cup B) = \frac{3}{4} \)
- \( P(A) = \frac{1}{3} \)

### Step-by-Step Solution:

1. **Understanding the Events**:
- Let \( A \) and \( B \) be two events.
- \( \overline{A} \) represents the complement of event \( A \), which includes all outcomes that are not in \( A \).

2. **Using the Formula**:
- We know that:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
- Rearranging this gives us:
\[ P(B) = P(A \cup B) - P(A) + P(A \cap B) \]

3. **Finding \( P(B) \)**:
- We need to express \( P(\overline{A} \cap B) \) in terms of known probabilities.
- We can use the relationship:
\[ P(B) = P(\overline{A} \cap B) + P(A \cap B) \]
- Thus, we can express \( P(\overline{A} \cap B) \) as:
\[ P(\overline{A} \cap B) = P(B) - P(A \cap B) \]

4. **Finding \( P(A \cap B) \)**:
- To find \( P(A \cap B) \), we can use the total probability:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
- Plugging in the known values:
\[ \frac{3}{4} = \frac{1}{3} + P(B) - P(A \cap B) \]

5. **Finding a Common Denominator**:
- To solve for \( P(B) \) and \( P(A \cap B) \), we can express \( \frac{1}{3} \) as \( \frac{4}{12} \) and \( \frac{3}{4} \) as \( \frac{9}{12} \):
\[ \frac{9}{12} = \frac{4}{12} + P(B) - P(A \cap B) \]
- Rearranging gives:
\[ P(B) - P(A \cap B) = \frac{9}{12} - \frac{4}{12} = \frac{5}{12} \]

6. **Substituting Back**:
- Now we can substitute back into our equation:
\[ P(\overline{A} \cap B) = P(B) - P(A \cap B) = \frac{5}{12} \]

7. **Final Answer**:
- Therefore, the probability \( P(\overline{A} \cap B) \) is:
\[ \boxed{\frac{5}{12}} \]