If P(B)`=(3)/(4)` P(A cap B cap overset(-)(c ))=(1)/(3)` and `P(overset(" "-)(A) cap B cap overset(-)(C ))=(1)/(3)`, then `P(B cap C)` is equal to

Options

A. `(1)/(12)`
B. `(1)/(6)`
C. `(1)/(15)`
D. `(1)/(15)`

Answer

To solve the problem, we will use the given probabilities and the properties of probability. Let's denote the events as follows:

- Let \( P(B) = \frac{3}{4} \)
- Let \( P(A \cap B \cap \overline{C}) = \frac{1}{3} \)
- Let \( P(\overline{A} \cap B \cap \overline{C}) = \frac{1}{3} \)

We need to find \( P(B \cap C) \).

### Step-by-step Solution:

1. **Understand the relationships**: We know that the total probability of event B can be expressed as:
\[ P(B) = P(A \cap B \cap C) + P(A \cap B \cap \overline{C}) + P(\overline{A} \cap B \cap C) + P(\overline{A} \cap B \cap \overline{C}) \]

2. **Substituting known values**: From the problem, we have:
- \( P(A \cap B \cap \overline{C}) = \frac{1}{3} \)
- \( P(\overline{A} \cap B \cap \overline{C}) = \frac{1}{3} \)

Let \( x = P(A \cap B \cap C) \) and \( y = P(\overline{A} \cap B \cap C) \).

Therefore, we can rewrite the equation as:
\[ P(B) = x + \frac{1}{3} + y + \frac{1}{3} \]
Simplifying gives:
\[ P(B) = x + y + \frac{2}{3} \]

3. **Substituting \( P(B) \)**: We know \( P(B) = \frac{3}{4} \), so we can set up the equation:
\[ \frac{3}{4} = x + y + \frac{2}{3} \]

4. **Finding a common denominator**: The common denominator for 4 and 3 is 12. We rewrite the equation:
\[ \frac{3}{4} = \frac{9}{12}, \quad \frac{2}{3} = \frac{8}{12} \]
Thus, we have:
\[ \frac{9}{12} = x + y + \frac{8}{12} \]

5. **Isolating \( x + y \)**:
\[ x + y = \frac{9}{12} - \frac{8}{12} = \frac{1}{12} \]

6. **Finding \( P(B \cap C) \)**: We know that:
\[ P(B \cap C) = P(A \cap B \cap C) + P(\overline{A} \cap B \cap C) = x + y \]
Therefore,
\[ P(B \cap C) = \frac{1}{12} \]

### Final Answer:
Thus, \( P(B \cap C) = \frac{1}{12} \).

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