If `A`, `B` and `C` are three exhaustive and mutually exclusive events such that `P(B) = 3/2P(A)` and `P( C) = 1/2P(B)`, then `P(AcupC)` is

Options

A. `3/13`
B. `6/13`
C. `7/13`
D. `10/13`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the probabilities
Let \( P(A) = x \). According to the problem, we have:
- \( P(B) = \frac{3}{2} P(A) = \frac{3}{2} x \)
- \( P(C) = \frac{1}{2} P(B) = \frac{1}{2} \cdot \frac{3}{2} x = \frac{3}{4} x \)

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

### Step 3: Combine the probabilities
To combine the terms, we need a common denominator. The least common multiple of the denominators (1, 2, and 4) is 4. Rewriting the equation:
\[ x + \frac{6}{4}x + \frac{3}{4}x = 1 \]
This simplifies to:
\[ x + \frac{9}{4}x = 1 \]
Combining the terms gives:
\[ \frac{4}{4}x + \frac{9}{4}x = \frac{13}{4}x = 1 \]

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

### Step 5: Find \( P(B) \) and \( P(C) \)
Now we can find \( P(B) \) and \( P(C) \):
\[ P(B) = \frac{3}{2} P(A) = \frac{3}{2} \cdot \frac{4}{13} = \frac{6}{13} \]
\[ P(C) = \frac{3}{4} P(A) = \frac{3}{4} \cdot \frac{4}{13} = \frac{3}{13} \]

### Step 6: Calculate \( P(A \cup C) \)
Since \( A \) and \( C \) are mutually exclusive:
\[ P(A \cup C) = P(A) + P(C) \]
Substituting the values:
\[ P(A \cup C) = \frac{4}{13} + \frac{3}{13} = \frac{7}{13} \]

### Final Answer
Thus, the probability \( P(A \cup C) \) is:
\[ \boxed{\frac{7}{13}} \]