If $A,B,C$ are three mutually exclusive and exhaustive events such that $P\left(A\right)=2P\left(B\right)=3P\left(C\right)$ then $P\left(A\right)=$   

$A\cup B\cup C=S,A\cap B=B\cap C=C\cap A=\varphi$

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

$P\left(A\right)=2P\left(B\right)=3P\left(C\right)=k$

$\Rightarrow k+\frac{k}{2}+\frac{k}{3}=1\Rightarrow k=\frac{6}{11}$

$\therefore P\left(A\right)=\frac{6}{11}$