Solution
= Solution
The <complement of an event> here is that all three dice avoid six, with <probability> $(5/6)^3$. Therefore
$$
\boxed{P(\text{at least one six})=1-(5/6)^3=\frac{91}{216}}.
$$
= Solution
The <complement of an event> here is that all three dice avoid six, with <probability> $(5/6)^3$. Therefore
$$
\boxed{P(\text{at least one six})=1-(5/6)^3=\frac{91}{216}}.
$$