If you roll a fair die six times, what is the probability of getting zero 6's?

Question

I know the probability of getting zero sixes on one roll is $5/6$, would I just find $(5/6)^6$ and subtract that from $1$ to get the probability?

Any help would be appreciated.

Answer 1

For each roll the probability is $\frac56$ thus for six rolls

$$P=\left(\frac56\right)^6$$

Note that $1-P$ is the probability of the complemantary case with at least one 6 in six rolls.

Answer 2

Assuming each dice roll is independent, and denoting the dice roll on the nth throw to be $X_n$, we have $$P(X_1\ne 6,X_2\ne6,\cdots,X_6\ne 6)=P(X_1\ne6)\times\cdots\times P(X_6\ne6)\\=P(X_1\ne6)^6\\=\left(\frac56\right)^6$$

You described subtracting this from $1$. This would give you the probability that at least one of the 6 dice rolls will be a 6. This is different to what you are looking for.