A fair die is rolled three times. What is the probability of getting at most one six?

'At most one' covers two cases. No six at all: $(5/6{)}^3=125/216$. Exactly one six: the six may land on any one of the three rolls, so $3\times (1/6)\times (5/6{)}^2=75/216$. Together that is 200/216 = 25/27. Stopping at the first case gives 125/216; 91/216 = 1 − 125/216 answers 'at least one six', the wrong complement; and 5/72 = 15/216 is the probability of exactly two sixes.