Probability of rolling three dice without getting a 6 I am having trouble understanding how you get $91/216$ as the answer to this question. 

say a die is rolled three times
what is the probability that at least one roll is 6?


Compute: 
$1 - \left(\dfrac{5}{6}\right)^3\; \;$ to obtain the probability that at least one 6 is rolled in three tosses.
That is, we find the probability that at least one $6$ is rolled by subtracting from $1$ the probability that it does not appear on any of three rolls:
The probability that $6$ does not appear given one roll is $5/6.\;$  The probability that a $6$ does not appear on three roles of the die is given by $$\left(\frac 56\right)^3 = \dfrac{125}{216}$$


*

*We use $1$ to represent certainty: that is 100 percent probability. It must be the case that P(Not rolling at least one 6 on three rolls) + P(rolling at least one 6 on three rolls)] =1$\implies 1 - $ (probability of not rolling at least one 6) = (probability of rolling at least one 6).


So the probability of obtaining a $6$ on at least one roll is:
$$1 - \dfrac{125}{216} \quad = \quad \dfrac{216}{216} - \dfrac{125}{216} \quad = \quad\dfrac{91}{216}$$
