If A, B, C are three mutually exclusive and exhaustive events of an experiment such that 4P(A) = 3P(B)=2P(C), then P(B) is equal to (i) `1/13` (ii) `2/13` (iii) `3/13` (iv) `4/13`

Options

A. `1/13`
B. `2/13`
C. `3/13`
D. `4/13`

Answer

To solve the problem, we will follow these steps:

### Step 1: Set up the equations based on the given relationships.
We know that:
1. \(4P(A) = 3P(B)\)
2. \(3P(B) = 2P(C)\)

### Step 2: Express \(P(A)\) and \(P(C)\) in terms of \(P(B)\).
From the first equation:
\[ P(A) = \frac{3}{4}P(B) \quad \text{(Equation 1)} \]
From the second equation:
\[ P(C) = \frac{3}{2}P(B) \quad \text{(Equation 2)} \]

### Step 3: Use the fact that the events are mutually exclusive and exhaustive.
Since A, B, and C are mutually exclusive and exhaustive events, we have:
\[ P(A) + P(B) + P(C) = 1 \]

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

### Step 5: Combine the terms.
To combine the terms, we need a common denominator. The least common multiple of 4 and 2 is 4. Thus, we rewrite the equation:
\[ \frac{3}{4}P(B) + \frac{4}{4}P(B) + \frac{6}{4}P(B) = 1 \]
This simplifies to:
\[ \frac{3 + 4 + 6}{4}P(B) = 1 \]
\[ \frac{13}{4}P(B) = 1 \]

### Step 6: Solve for \(P(B)\).
Now, we can solve for \(P(B)\):
\[ P(B) = \frac{4}{13} \]

### Conclusion:
Thus, the probability \(P(B)\) is equal to \(\frac{4}{13}\).