If for two events A and B, in a random experiment, P(A|B) = 4/5 and P(B|A) = 1/4, then P(A | A `uu` B) is equal to _____

Options

A. `5//16`
B. `5//17`
C. `16//17`
D. `11//16`

Answer

To solve the problem, we need to find \( P(A | A \cup B) \) given the probabilities \( P(A|B) = \frac{4}{5} \) and \( P(B|A) = \frac{1}{4} \).

### Step-by-Step Solution:

1. **Understanding the Formula**:
The conditional probability can be expressed as:
\[ P(A | A \cup B) = \frac{P(A \cap (A \cup B))}{P(A \cup B)} \]
Since \( A \cap (A \cup B) = A \), we can simplify this to:
\[ P(A | A \cup B) = \frac{P(A)}{P(A \cup B)} \]

2. **Finding \( P(A \cup B) \)**:
We use the formula for the union of two events:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

3. **Finding \( P(A \cap B) \)**:
We can use the definitions of conditional probabilities:
- From \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), we have:
\[ P(A \cap B) = P(A|B) \cdot P(B) = \frac{4}{5} P(B) \]
- From \( P(B|A) = \frac{P(A \cap B)}{P(A)} \), we have:
\[ P(A \cap B) = P(B|A) \cdot P(A) = \frac{1}{4} P(A) \]

Setting these two expressions for \( P(A \cap B) \) equal gives:
\[ \frac{4}{5} P(B) = \frac{1}{4} P(A) \]

4. **Solving for \( P(B) \)**:
Rearranging gives:
\[ P(B) = \frac{1}{4} \cdot \frac{5}{4} P(A) = \frac{5}{16} P(A) \]

5. **Substituting Back**:
Now we can substitute \( P(B) \) back into the equation for \( P(A \cup B) \):
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
We already have \( P(A \cap B) = \frac{4}{5} P(B) = \frac{4}{5} \cdot \frac{5}{16} P(A) = \frac{1}{4} P(A) \).

Thus,
\[ P(A \cup B) = P(A) + \frac{5}{16} P(A) - \frac{1}{4} P(A) \]
Converting \( \frac{1}{4} P(A) \) to sixteenths gives \( \frac{4}{16} P(A) \):
\[ P(A \cup B) = P(A) + \frac{5}{16} P(A) - \frac{4}{16} P(A) = P(A) + \frac{1}{16} P(A) = \frac{17}{16} P(A) \]

6. **Final Calculation**:
Now substituting back into the conditional probability:
\[ P(A | A \cup B) = \frac{P(A)}{P(A \cup B)} = \frac{P(A)}{\frac{17}{16} P(A)} = \frac{16}{17} \]

Thus, the final answer is:
\[ P(A | A \cup B) = \frac{16}{17} \]