Question

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

Answer 1

Let the probability of winning by C i.e., P(C) be $x$

$\therefore P\left(B\right)=2x$ and

$P\left(A\right)=2\left(2x\right)=4x$

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

$4x+2x+x=1$

$7x=1$

$x=\frac{1}{7}$

Hence $P\left(A\right)=\frac{4}{7};P\left(B\right)=\frac{2}{7}$ and $P\left(C\right)=\frac{1}{7}$