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:

$$1 - \dfrac{125}{216} \quad = \quad \dfrac{216}{216} - \dfrac{125}{216} \quad = \quad\dfrac{91}{216}$$
For rolling 3 dice without getting a six, to get the probability, pretend your 6-sided dice are 5-sided dice.  That can help you figure out the probability.  Therefore, the probability is 125/216.