Let A, B, C be three events such that P (exactly one of A or (B) = P (exactly one of B or ( C ) = P (exactly one of C or (A) = `(1)/(3)` and P `(A nn B nn C)=(1)/(9)`, then `P(A uu B uu C)` is equal to

Options

A. `11//18`
B. `9//11`
C. `1//2`
D. `11//36`

Answer

To solve the problem, we need to find the probability \( P(A \cup B \cup C) \) given the conditions about the probabilities of events A, B, and C.

### Step-by-step Solution:

1. **Understanding the Given Information:**
We know:
- \( P(\text{exactly one of } A \text{ or } B) = \frac{1}{3} \)
- \( P(\text{exactly one of } B \text{ or } C) = \frac{1}{3} \)
- \( P(\text{exactly one of } C \text{ or } A) = \frac{1}{3} \)
- \( P(A \cap B \cap C) = \frac{1}{9} \)

2. **Setting Up the Equations:**
We can express the probabilities of exactly one of the events using the formula:
\[ P(\text{exactly one of } A \text{ or } B) = P(A) + P(B) - 2P(A \cap B) \]
Therefore, we have:
\[ P(A) + P(B) - 2P(A \cap B) = \frac{1}{3} \quad \text{(1)} \]
Similarly, for the other pairs:
\[ P(B) + P(C) - 2P(B \cap C) = \frac{1}{3} \quad \text{(2)} \]
\[ P(C) + P(A) - 2P(C \cap A) = \frac{1}{3} \quad \text{(3)} \]

3. **Adding the Equations:**
Adding equations (1), (2), and (3):
\[ (P(A) + P(B) - 2P(A \cap B)) + (P(B) + P(C) - 2P(B \cap C)) + (P(C) + P(A) - 2P(C \cap A)) = 1 \]
Simplifying this gives:
\[ 2P(A) + 2P(B) + 2P(C) - 2(P(A \cap B) + P(B \cap C) + P(C \cap A)) = 1 \]
Dividing the entire equation by 2:
\[ P(A) + P(B) + P(C) - (P(A \cap B) + P(B \cap C) + P(C \cap A)) = \frac{1}{2} \quad \text{(4)} \]

4. **Finding \( P(A \cup B \cup C) \):**
The formula for the probability of the union of three events is:
\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - (P(A \cap B) + P(B \cap C) + P(C \cap A)) + P(A \cap B \cap C) \]
Substituting from equation (4) and the given \( P(A \cap B \cap C) = \frac{1}{9} \):
\[ P(A \cup B \cup C) = \frac{1}{2} + \frac{1}{9} \]

5. **Calculating the Final Probability:**
To add \( \frac{1}{2} \) and \( \frac{1}{9} \), we need a common denominator:
\[ \frac{1}{2} = \frac{9}{18}, \quad \frac{1}{9} = \frac{2}{18} \]
Therefore,
\[ P(A \cup B \cup C) = \frac{9}{18} + \frac{2}{18} = \frac{11}{18} \]

### Final Answer:
\[ P(A \cup B \cup C) = \frac{11}{18} \]