Question

A random experiment has only three pairs of mutually exclusive events among $A,B$ and $C$. If $P\left(A\right)=\frac{3}{2},P\left(B\right)$ and $P\left(C\right)=\frac{1}{2}P\left(A\right)$, find $P\left(B\right)$

Answer 1

Let $P\left(B\right)=a$
Now, $P\left(A\right)=\frac{3}{2}P\left(B\right)=\frac{3}{2}a$
Also, $P\left(C\right)=\frac{1}{2}P\left(A\right)=\frac{1}{2}\times \frac{3}{2}a=\frac{3}{4}a$
It is given that $A,B$ and $C$ are mutually exclusive events.
$\therefore P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)$
Since $P\left(S\right)=1$, we get $P\left(A\right)+P\left(B\right)+P\left(C\right)=1$
$\therefore \frac{3}{2}a+a+\frac{3}{4}a=1,\frac{6a+4a+3a}{4}=1,\therefore 13a=4$
$\therefore a=\frac{4}{13}$ or $P\left(B\right)=\frac{4}{13}$