If `A, B` are mutually exclusive events such that `P(A cup B) = 5/6 and P(B) = 1/3," then " P(A')=`

Options

A. `1/2`
B. `1/3`
C. `1/4`
D. `5/18`

Answer

To solve the problem, we need to find \( P(A') \) given that \( P(A \cup B) = \frac{5}{6} \) and \( P(B) = \frac{1}{3} \).

### Step-by-step Solution:

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

2. **Use the Formula for Union of Two Events**:
The probability of the union of two events \( A \) and \( B \) is given by:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Since \( P(A \cap B) = 0 \), the formula simplifies to:
\[ P(A \cup B) = P(A) + P(B) \]

3. **Substitute the Known Values**:
We know:
\[ P(A \cup B) = \frac{5}{6} \quad \text{and} \quad P(B) = \frac{1}{3} \]
Substituting these values into the equation gives:
\[ \frac{5}{6} = P(A) + \frac{1}{3} \]

4. **Convert \( P(B) \) to a Common Denominator**:
To combine the fractions, we convert \( \frac{1}{3} \) to have a denominator of 6:
\[ \frac{1}{3} = \frac{2}{6} \]
Now, substitute this back into the equation:
\[ \frac{5}{6} = P(A) + \frac{2}{6} \]

5. **Solve for \( P(A) \)**:
Rearranging the equation to isolate \( P(A) \):
\[ P(A) = \frac{5}{6} - \frac{2}{6} = \frac{3}{6} = \frac{1}{2} \]

6. **Find \( P(A') \)**:
The probability of the complement of event \( A \), denoted as \( P(A') \), is given by:
\[ P(A') = 1 - P(A) \]
Substituting the value of \( P(A) \):
\[ P(A') = 1 - \frac{1}{2} = \frac{1}{2} \]

### Final Answer:
Thus, the probability \( P(A') \) is \( \frac{1}{2} \).

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