Question

In a certain race there are three boys $A,B,C$. The winning probability of $A$ is twice than $B$ and the winning probability of $B$ is twice than $C$. If $P\left(A\right)+P\left(B\right)+P\left(C\right)=1$, then find the probability of each boy.

Answer 1

Probability of $A$ wins i.e. $P\left(A\right)=2P\left(B\right)$ and $P\left(B\right)=2P\left(C\right)$
$3P\left(B\right)+\frac{1}{2}P\left(B\right)=1$
$\frac{7}{2}P\left(B\right)=1$
$\therefore P\left(B\right)=\frac{2}{7}$
Now, $P\left(A\right)=2P\left(B\right)$
$=2\times \frac{2}{7}$
$=\frac{4}{7}$
and $P\left(C\right)=\frac{P\left(B\right)}{2}$
$=\frac{1}{2}\times \frac{2}{7}$
$=\frac{1}{7}$
Hence, $P\left(A\right)=\frac{4}{7}$
$P\left(B\right)=\frac{2}{7}$
and $P\left(C\right)=\frac{1}{7}$