Question

A, B, C are three mutually exclusive and exhaustive events associated with a random experiment. Find $P\left(A\right)$, it being given that $P\left(B\right)=\frac{3}{2}P\left(A\right)$ and $P\left(C\right)=\frac{1}{2}P\left(B\right)$.

• A. $\frac{1}{16}$
• B. $\frac{16}{81}$
• C. $\frac{4}{13}$
• D. None of these

Answer 1

The correct option is C $\frac{4}{13}$

Let $P\left(A\right)=p$. Then,

$P\left(B\right)=\frac{3}{2}P\left(A\right)⟹P\left(B\right)=\frac{3}{2}p$ and $P\left(C\right)=\frac{1}{2}P\left(B\right)⟹P\left(C\right)=\frac{3}{4}p$

Since A, B, C are mutually exclusive and exhaustive events associated with a random experiment.

$A\cup B\cup C=S$

$P(A\cup B\cup C)=P\left(S\right)$

$P(A\cup B\cup C)=1$

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

$p+\frac{3}{2}p+\frac{3}{4}p=1$

$p=\frac{4}{13}$

$P\left(A\right)=\frac{4}{13}$