If three dice are rolled under the condition that no two dice show...

Number of ways in which one of the face having the number 6 and no two dice show the same number. $$\begin{aligned} & (1,2,6),(1,3,6),(1,4,6),(1,5,6),(2,3,6),(2,4,6),(2,5,6),(3,4, \\ & 6),(3,5,6) \ldots \ldots . \\ & \text{ Total favourable case }=20+20+20=60 \end{aligned}$$ Number of total output when three top faces of three dice shows different number $=6 \times 5 \times 4=120$. $\therefore$ Required probability =$\frac{60}{120}=\frac{1}{2}.$