Rolling $20$ dice at least three six

Question

If we roll $20$ dice, what is the probability of at least three six?

I solved by using $P(A)=1-P(A^C)$ property. Considering the complement of problem, there is a $5/6$ probability of not rolling a six for any given die. So the result is $1-{\left(\frac{5}{6}\right)}^{20}$ Am I thinking wrong? Any help will be appreciated.

Answer 1

$$1-\binom{20}{0}\Big(\frac{1}{6}\Big)^0\Big(\frac{5}{6}\Big)^{20}-\binom{20}{1}\Big(\frac{1}{6}\Big)^1\Big(\frac{5}{6}\Big)^{19}-\binom{20}{2}\Big(\frac{1}{6}\Big)^2\Big(\frac{5}{6}\Big)^{18}$$

That is

$$1-\text{Probability of zero six's}-\text{Probability of 1 six}-\text{Probability of 2 six's}$$