In a competition $A, B, C$ are participating the probability that $A$ wins is twice that of $B$, the…
In a competition $A, B, C$ are participating the probability that $A$ wins is twice that of $B$, the probability that $B$ wins is twice that of $C$, then probability that $A$ loses is
$\frac{1}{7}$
$\frac{2}{7}$
$\frac{4}{7}$
$\frac{3}{7}$
Solution
Let $P(C)=p$, then $P(B)=2 p, P(A)=2(2 p)=4 p$
Given that, $P(A)+P(B)+P(C)=1$
$\begin{aligned} \Rightarrow & & 4 p+2 p+p+1 \\ \Rightarrow & & p=\frac{1}{7} \\ & \therefore & P(A)=\frac{4}{7} \\ & & P(\bar{A})=1-\frac{4}{7}=\frac{3}{7}\end{aligned}$
Thus, the required probability is $\frac{3}{7}$.