Question

A,B and C are three mutually exclusive and exhaustive events and $P\left(B\right)=\frac{3}{2}P\left(A\right),P\left(C\right)=\frac{1}{3}P\left(B\right)$ then the value of $P\left(A\right)$ is

• A. 1/3
• B. 1/2
• C. 1/6
• D. none of these

Answer 1

Answer: (A)

1/3

Given, $A,B$ and $C$ are three mutually exclusive and exhaustive events and $P\left(B\right)=\frac{3}{2}P\left(A\right),P\left(C\right)=\frac{1}{3}P\left(B\right)$.

Since $A,B,C$ are mutually exclusive and exhaustive events then we've,

$P\left(A\right)+P\left(B\right)+P\left(C\right)=1$

or, $P\left(A\right)+\frac{3}{2}P\left(A\right)+\frac{1}{2}P\left(A\right)=1$

or, $3P\left(A\right)=1$

or, $P\left(A\right)=\frac{1}{3}$.