Question

If \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) are mutually exclusive and exhaustive events of a sample space \(S\) such that \(P(B)=\frac{3}{2} P(A)\) and \(P(C)=\frac{1}{2} P(B)\), then \(P(A)=\)

• A. \(\frac{4}{13}\)
• B. \(\frac{6}{13}\)
• C. \(\frac{8}{13}\)
• D. \(\frac{3}{13}\)

Hint:

Express all three probabilities through P(A) and use the fact that they add up to 1.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For mutually exclusive and exhaustive events, \(P(A)+P(B)+P(C)=1\).

Step 2: Express through P(A)
\(P(B)=\dfrac{3}{2}P(A)\)
\(P(C)=\dfrac{1}{2}P(B)=\dfrac{3}{4}P(A)\)

Step 3: Calculation
\(P(A)\left(1+\dfrac{3}{2}+\dfrac{3}{4}\right)=1\Rightarrow P(A)\cdot\dfrac{13}{4}=1\)

Step 4: Result
\(P(A)=\dfrac{4}{13}\), the first option.

Final Answer:
The probability of A is 4/13. \[ \boxed{\text{(A) }\dfrac{4}{13}} \]