Probability of winning of $A$ $=P\left(A\right)=\frac{4}{6}=\frac{2}{3}$

$\Rightarrow P\left(\overline{A}\right)=1-P\left(A\right)=\frac{1}{3}$

Probability of winning of $B$ =$P\left(B\right)=\frac{3}{5}$

$\Rightarrow P\left(\overline{B}\right)=1-P\left(B\right)=\frac{2}{5}$

Probability of winning of $C$ $=P\left(C\right)=\frac{2}{3}$

$\Rightarrow P\left(\overline{C}\right)=1-P\left(C\right)=\frac{1}{3}$

Probability of winning atleast means one player will loose $=P\left(\overline{A}\right)P\left(B\right)P\left(C\right)+P\left(A\right)P\left(\overline{B}\right)P\left(C\right)+P\left(A\right)P\left(B\right)P\left(\overline{C}\right)$

$=\frac{1}{3}\times \frac{3}{5}\times \frac{2}{3}+\frac{2}{3}\times \frac{2}{5}\times \frac{2}{3}+\frac{2}{3}\times \frac{3}{5}\times \frac{1}{3}$

$=\frac{4}{9}$