If A,B and C are mutually exclusive and exhaustive events of a random experiment such that `P(B)=(3)/(2)P(A)` and `P(C )=(1)/(2)P(B)`, then `P (A uu C)` equals

Options

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

Explanation

To solve the problem, we need to find \( P(A \cup C) \) given the relationships between the probabilities of events A, B, and C. Let's break it down step by step.

### Step 1: Understand the relationships
We are given:
1. \( P(B) = \frac{3}{2} P(A) \)
2. \( P(C) = \frac{1}{2} P(B) \)

Since A, B, and C are mutually exclusive and exhaustive events, we know:
\[ P(A) + P(B) + P(C) = 1 \]

### Step 2: Express \( P(B) \) and \( P(C) \) in terms of \( P(A) \)
From the first equation:
\[ P(B) = \frac{3}{2} P(A) \]

From the second equation, substituting \( P(B) \):
\[ P(C) = \frac{1}{2} P(B) = \frac{1}{2} \left(\frac{3}{2} P(A)\right) = \frac{3}{4} P(A) \]

### Step 3: Substitute into the total probability equation
Now substitute \( P(B) \) and \( P(C) \) into the total probability equation:
\[ P(A) + P(B) + P(C) = 1 \]
\[ P(A) + \frac{3}{2} P(A) + \frac{3}{4} P(A) = 1 \]

### Step 4: Combine the terms
To combine these terms, we need a common denominator. The least common multiple of 1, 2, and 4 is 4. Therefore, we rewrite the equation:
\[ P(A) = \frac{4}{4} P(A) \]
\[ \frac{4}{4} P(A) + \frac{6}{4} P(A) + \frac{3}{4} P(A) = 1 \]
\[ \frac{4 + 6 + 3}{4} P(A) = 1 \]
\[ \frac{13}{4} P(A) = 1 \]

### Step 5: Solve for \( P(A) \)
Now, we can solve for \( P(A) \):
\[ P(A) = \frac{4}{13} \]

### Step 6: Find \( P(B) \) and \( P(C) \)
Now that we have \( P(A) \), we can find \( P(B) \) and \( P(C) \):
\[ P(B) = \frac{3}{2} P(A) = \frac{3}{2} \times \frac{4}{13} = \frac{12}{13} \]
\[ P(C) = \frac{3}{4} P(A) = \frac{3}{4} \times \frac{4}{13} = \frac{3}{13} \]

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

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