When three standard dice are tossed, the numbers $a,b,c$ are obtained. Find the probability that $abc = 180$.
We first factor 180 into three positive integers from the set $\{1,2,3,4,5,6\}$. Since $180 > 5^3 = 125,$ at least one of the integers must be 6. Since $180 > 5^2\cdot 6 = 150$, at least two integers must equal 6. Indeed, $180 = 5\cdot6\cdot6$ is the only such way to factor 180. Therefore, $(a,b,c) = (5,6,6), (6,5,6),(6,6,5)$ are the only possibilities for $a,b,c$. Each occurs with probability $\left(\frac16\right)^3 = \frac1{216}$, so the probability that $abc = 180$ is $3\cdot \frac1{216} = \boxed{\frac1{72}}$.