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)/(16)`
D. `(1)/(9)`

Explanation

To solve the problem, we need to find \( P(B \cap C) \) given the following probabilities:

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

### Step 1: Understand the given probabilities

We have:
- \( P(B) = \frac{3}{4} \) means that the probability of event B occurring is \( \frac{3}{4} \).
- \( P(A \cap B \cap \overline{C}) = \frac{1}{3} \) means that the probability of both A and B occurring while C does not occur is \( \frac{1}{3} \).
- \( P(\overline{A} \cap B \cap \overline{C}) = \frac{1}{3} \) means that the probability of B occurring while A does not occur and C does not occur is also \( \frac{1}{3} \).

### Step 2: Calculate \( P(B \cap \overline{C}) \)

To find \( P(B \cap \overline{C}) \), we can add the probabilities of the two events involving B and \( \overline{C} \):

\[ P(B \cap \overline{C}) = P(A \cap B \cap \overline{C}) + P(\overline{A} \cap B \cap \overline{C}) \]

Substituting the values we have:

\[ P(B \cap \overline{C}) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \]

### Step 3: Use the total probability of B

We know that:

\[ P(B) = P(B \cap C) + P(B \cap \overline{C}) \]

Substituting the known values:

\[ \frac{3}{4} = P(B \cap C) + \frac{2}{3} \]

### Step 4: Solve for \( P(B \cap C) \)

To isolate \( P(B \cap C) \), we can rearrange the equation:

\[ P(B \cap C) = \frac{3}{4} - \frac{2}{3} \]

To perform this subtraction, we need a common denominator. The least common multiple of 4 and 3 is 12.

Convert \( \frac{3}{4} \) and \( \frac{2}{3} \) to have a denominator of 12:

\[ \frac{3}{4} = \frac{9}{12} \]
\[ \frac{2}{3} = \frac{8}{12} \]

Now substitute back into the equation:

\[ P(B \cap C) = \frac{9}{12} - \frac{8}{12} = \frac{1}{12} \]

### Conclusion

Thus, the probability \( P(B \cap C) \) is:

\[ \boxed{\frac{1}{12}} \]