Suppose A, B and C are three mutually exclusive and exhaustive events such that 3P(A) = 2P(B) and P(B) = 2P(C), then `1/(P(A uu B))`=_____

Answer

To solve the problem step by step, we will use the given relationships between the probabilities of events A, B, and C, and the fact that they are mutually exclusive and exhaustive.

### Step 1: Define the Probabilities
Let:
- \( P(A) = x \)
- \( P(B) = y \)
- \( P(C) = z \)

From the problem, we have the following relationships:
1. \( 3P(A) = 2P(B) \) ⇒ \( 3x = 2y \) ⇒ \( y = \frac{3}{2}x \)
2. \( P(B) = 2P(C) \) ⇒ \( y = 2z \) ⇒ \( z = \frac{y}{2} \)

### Step 2: Substitute \( y \) in terms of \( x \)
Using \( y = \frac{3}{2}x \) in the equation for \( z \):
\[ z = \frac{y}{2} = \frac{\frac{3}{2}x}{2} = \frac{3}{4}x \]

### Step 3: Use the Exhaustive Condition
Since A, B, and C are mutually exclusive and exhaustive, we have:
\[ P(A) + P(B) + P(C) = 1 \]
Substituting the values of \( P(A) \), \( P(B) \), and \( P(C) \):
\[ x + \frac{3}{2}x + \frac{3}{4}x = 1 \]

### Step 4: Find a Common Denominator
The common denominator for the fractions is 4. Rewriting the equation:
\[ \frac{4}{4}x + \frac{6}{4}x + \frac{3}{4}x = 1 \]
Combining the terms:
\[ \frac{4 + 6 + 3}{4}x = 1 \Rightarrow \frac{13}{4}x = 1 \]

### Step 5: Solve for \( x \)
Now, solving for \( x \):
\[ x = \frac{4}{13} \]
Thus, \( P(A) = \frac{4}{13} \).

### Step 6: Find \( P(B) \) and \( P(C) \)
Using \( y = \frac{3}{2}x \):
\[ P(B) = y = \frac{3}{2} \cdot \frac{4}{13} = \frac{6}{13} \]
Using \( z = \frac{3}{4}x \):
\[ P(C) = z = \frac{3}{4} \cdot \frac{4}{13} = \frac{3}{13} \]

### Step 7: Find \( P(A \cup B) \)
Since A and B are mutually exclusive:
\[ P(A \cup B) = P(A) + P(B) = \frac{4}{13} + \frac{6}{13} = \frac{10}{13} \]

### Step 8: Find \( \frac{1}{P(A \cup B)} \)
Finally, we need to find:
\[ \frac{1}{P(A \cup B)} = \frac{1}{\frac{10}{13}} = \frac{13}{10} \]

### Final Answer
Thus, the value of \( \frac{1}{P(A \cup B)} \) is:
\[ \boxed{\frac{13}{10}} \]