Let A and B are two mutually exclusive events with `P(A)= (1)/(3)` and `P(B)= (1)/(4)` What is the value of P `(barA nn bar B)` ?

Options

A. `(1)/(6)`
B. `(1)/(4)`
C. `(1)/(3)`
D. `(5)/(12)`

Answer

To solve the problem, we need to find the value of \( P(\bar{A} \cap \bar{B}) \), where \( A \) and \( B \) are mutually exclusive events with given probabilities.

### Step-by-Step Solution:

1. **Understanding Mutually Exclusive Events**:
- Since \( A \) and \( B \) are mutually exclusive, it means that both events cannot occur at the same time. Therefore, \( P(A \cap B) = 0 \).

2. **Given Probabilities**:
- We are given \( P(A) = \frac{1}{3} \) and \( P(B) = \frac{1}{4} \).

3. **Finding \( P(A \cup B) \)**:
- The formula for the probability of the union of two events is:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
- Substituting the values we have:
\[ P(A \cup B) = P(A) + P(B) - 0 = \frac{1}{3} + \frac{1}{4} \]

4. **Finding a Common Denominator**:
- The least common multiple (LCM) of 3 and 4 is 12. We convert the fractions:
\[ P(A) = \frac{1}{3} = \frac{4}{12}, \quad P(B) = \frac{1}{4} = \frac{3}{12} \]
- Now, we can add them:
\[ P(A \cup B) = \frac{4}{12} + \frac{3}{12} = \frac{7}{12} \]

5. **Finding \( P(\bar{A} \cap \bar{B}) \)**:
- The probability of the complement of the union of two events is given by:
\[ P(\bar{A} \cap \bar{B}) = 1 - P(A \cup B) \]
- Substituting the value we found:
\[ P(\bar{A} \cap \bar{B}) = 1 - \frac{7}{12} \]
- To perform the subtraction:
\[ P(\bar{A} \cap \bar{B}) = \frac{12}{12} - \frac{7}{12} = \frac{5}{12} \]

6. **Final Answer**:
- Therefore, the value of \( P(\bar{A} \cap \bar{B}) \) is \( \frac{5}{12} \).