Let A and B are two independent events such that `P(B)=(1)/(2)` and `P(AnnB)=(1)/(10)`, then the value of `9P((barA)/(AuuB))` is

Answer

To solve the problem, we need to find the value of \( 9P(\bar{A} | A \cup B) \) given that \( P(B) = \frac{1}{2} \) and \( P(A \cap B) = \frac{1}{10} \).

### Step-by-Step Solution:

1. **Identify the Known Values:**
- \( P(B) = \frac{1}{2} \)
- \( P(A \cap B) = \frac{1}{10} \)

2. **Use the Independence of Events A and B:**
Since A and B are independent events, we can use the formula:
\[ P(A \cap B) = P(A) \cdot P(B) \]
Substituting the known values:
\[ \frac{1}{10} = P(A) \cdot \frac{1}{2} \]

3. **Solve for \( P(A) \):**
Rearranging the equation gives:
\[ P(A) = \frac{1}{10} \cdot 2 = \frac{2}{10} = \frac{1}{5} \]

4. **Calculate \( P(A \cup B) \):**
We use the formula for the union of two events:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting the values we found:
\[ P(A \cup B) = \frac{1}{5} + \frac{1}{2} - \frac{1}{10} \]
To add these fractions, we need a common denominator (which is 10):
\[ P(A \cup B) = \frac{2}{10} + \frac{5}{10} - \frac{1}{10} = \frac{6}{10} = \frac{3}{5} \]

5. **Calculate \( P(\bar{A}) \):**
We know that:
\[ P(\bar{A}) = 1 - P(A) = 1 - \frac{1}{5} = \frac{4}{5} \]

6. **Calculate \( P(\bar{A} | A \cup B) \):**
Using the definition of conditional probability:
\[ P(\bar{A} | A \cup B) = \frac{P(\bar{A} \cap (A \cup B))}{P(A \cup B)} \]
Here, \( P(\bar{A} \cap (A \cup B)) \) can be simplified as:
\[ P(\bar{A} \cap (A \cup B)) = P(B) - P(A \cap B) = \frac{1}{2} - \frac{1}{10} \]
Converting to a common denominator (10):
\[ P(B) - P(A \cap B) = \frac{5}{10} - \frac{1}{10} = \frac{4}{10} = \frac{2}{5} \]
Thus, we have:
\[ P(\bar{A} | A \cup B) = \frac{\frac{2}{5}}{\frac{3}{5}} = \frac{2}{3} \]

7. **Calculate \( 9P(\bar{A} | A \cup B) \):**
Finally, we compute:
\[ 9P(\bar{A} | A \cup B) = 9 \cdot \frac{2}{3} = 6 \]

### Final Answer:
The value of \( 9P(\bar{A} | A \cup B) \) is \( \boxed{6} \).