Let A, B and C be three mutually exclusive and exhaustive events associated with a random experiment. IfP(B) = 1.5 P(A) and P(C)=0.5P(B), then P(A) is equal to

Options

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

Explanation

To solve the problem step by step, we will use the given relationships between the probabilities of events A, B, and C.

### Step 1: Set up the equations based on the problem statement.
We know that:
1. \( P(B) = 1.5 P(A) \) (Equation 1)
2. \( P(C) = 0.5 P(B) \) (Equation 2)
3. Since A, B, and C are mutually exclusive and exhaustive events, we have:
\[ P(A) + P(B) + P(C) = 1 \quad (Equation 3) \]

### Step 2: Substitute \( P(B) \) and \( P(C) \) in terms of \( P(A) \).
From Equation 1, we can express \( P(B) \) in terms of \( P(A) \):
\[ P(B) = 1.5 P(A) \]

Now, substituting \( P(B) \) into Equation 2:
\[ P(C) = 0.5 P(B) = 0.5 \times 1.5 P(A) = 0.75 P(A) \]

### Step 3: Substitute \( P(B) \) and \( P(C) \) into Equation 3.
Now we can substitute \( P(B) \) and \( P(C) \) into Equation 3:
\[ P(A) + P(B) + P(C) = 1 \]
Substituting the values we found:
\[ P(A) + 1.5 P(A) + 0.75 P(A) = 1 \]

### Step 4: Combine like terms.
Combine the terms on the left side:
\[ P(A) + 1.5 P(A) + 0.75 P(A) = (1 + 1.5 + 0.75) P(A) = 3.25 P(A) \]
So we have:
\[ 3.25 P(A) = 1 \]

### Step 5: Solve for \( P(A) \).
Now, solve for \( P(A) \):
\[ P(A) = \frac{1}{3.25} = \frac{1}{\frac{13}{4}} = \frac{4}{13} \]

### Conclusion:
Thus, the probability of event A is:
\[ P(A) = \frac{4}{13} \]