If `P(A)=1//3, P(B) =3//4` and `P(A uu B)=11//12`, then what is `P(B//A)`?

Options

A. `1//6`
B. `4//9`
C. `1//2`
D. `1//3`

Answer

To solve the problem, we need to find the conditional probability \( P(B|A) \) given the probabilities \( P(A) = \frac{1}{3} \), \( P(B) = \frac{3}{4} \), and \( P(A \cup B) = \frac{11}{12} \).

### Step-by-Step Solution:

1. **Understanding the Formula for Conditional Probability**:
The formula for conditional probability is given by:
\[ P(B|A) = \frac{P(A \cap B)}{P(A)} \]
where \( P(A \cap B) \) is the probability of both events A and B occurring.

2. **Finding \( P(A \cap B) \)**:
We can use the formula for the probability of the union of two events:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Plugging in the values we have:
\[ \frac{11}{12} = \frac{1}{3} + \frac{3}{4} - P(A \cap B) \]

3. **Calculating the Left Side**:
First, we need to find a common denominator to add \( \frac{1}{3} \) and \( \frac{3}{4} \). The least common multiple of 3 and 4 is 12.
- Convert \( \frac{1}{3} \) to twelfths:
\[ \frac{1}{3} = \frac{4}{12} \]
- Convert \( \frac{3}{4} \) to twelfths:
\[ \frac{3}{4} = \frac{9}{12} \]

4. **Substituting Back**:
Now substitute these values back into the equation:
\[ \frac{11}{12} = \frac{4}{12} + \frac{9}{12} - P(A \cap B) \]
Simplifying the left side:
\[ \frac{11}{12} = \frac{13}{12} - P(A \cap B) \]

5. **Isolating \( P(A \cap B) \)**:
Rearranging the equation gives:
\[ P(A \cap B) = \frac{13}{12} - \frac{11}{12} = \frac{2}{12} = \frac{1}{6} \]

6. **Calculating \( P(B|A) \)**:
Now that we have \( P(A \cap B) \), we can find \( P(B|A) \):
\[ P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{6}}{\frac{1}{3}} \]

7. **Simplifying the Fraction**:
Dividing \( \frac{1}{6} \) by \( \frac{1}{3} \) is the same as multiplying by the reciprocal:
\[ P(B|A) = \frac{1}{6} \times \frac{3}{1} = \frac{3}{6} = \frac{1}{2} \]

### Final Answer:
Thus, the conditional probability \( P(B|A) \) is:
\[ \boxed{\frac{1}{2}} \]