Let A, B, C be three events and `barA, barB, barC` be their corresponding complementary event. If the probabilities of events `B, AnnBnnbarC and barA nnB nnbarC` are `(5)/(6),(1)/(2) and (1)/(4)` respectively, then the probability of the event `BnnC` is

Options

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

Answer

To find the probability of the event \( B \cap C \), we can use the information given in the problem.

### Step-by-Step Solution:

1. **Identify Given Probabilities**:
- \( P(B) = \frac{5}{6} \)
- \( P(A \cap B \cap \bar{C}) = \frac{1}{2} \)
- \( P(\bar{A} \cap B \cap \bar{C}) = \frac{1}{4} \)

2. **Understand the Components of \( P(B) \)**:
The event \( B \) can be broken down into three parts:
- \( A \cap B \cap C \) (Part of \( B \) where both \( A \) and \( C \) occur)
- \( A \cap B \cap \bar{C} \) (Part of \( B \) where \( A \) occurs but not \( C \))
- \( \bar{A} \cap B \cap \bar{C} \) (Part of \( B \) where neither \( A \) nor \( C \) occurs)

Hence, we can express \( P(B) \) as:
\[ P(B) = P(A \cap B \cap C) + P(A \cap B \cap \bar{C}) + P(\bar{A} \cap B \cap \bar{C}) \]

3. **Set Up the Equation**:
We can rearrange the equation to find \( P(B \cap C) \):
\[ P(B \cap C) = P(B) - P(A \cap B \cap \bar{C}) - P(\bar{A} \cap B \cap \bar{C}) \]

4. **Substitute the Known Values**:
Substitute the known probabilities into the equation:
\[ P(B \cap C) = \frac{5}{6} - \frac{1}{2} - \frac{1}{4} \]

5. **Finding a Common Denominator**:
The common denominator for \( 6, 2, \) and \( 4 \) is \( 12 \). Convert each fraction:
- \( \frac{5}{6} = \frac{10}{12} \)
- \( \frac{1}{2} = \frac{6}{12} \)
- \( \frac{1}{4} = \frac{3}{12} \)

6. **Perform the Calculation**:
Now substitute these values back into the equation:
\[ P(B \cap C) = \frac{10}{12} - \frac{6}{12} - \frac{3}{12} \]
\[ P(B \cap C) = \frac{10 - 6 - 3}{12} = \frac{1}{12} \]

### Final Answer:
The probability of the event \( B \cap C \) is \( \frac{1}{12} \).