Probability of rolling a fair dice m times to get a specific number n times.

I am tasked with the following series of questions.

Find the probability of each of the following:

1. Rolling 6 dice and getting at least one 3.


2. Rolling 12 dice and getting at least two 3s.


3. Rolling 18 dice and getting at least three 3s.

I am able to solve the first one. $P(A) = (1/6) + (1/6)(5/6) +... (1/6)(5/6)^5$. I then apply the geometric progression formula.
However, I can't seem to find a good way to solve questions 2 and 3 even though they look similar to the first one. I am probably missing some pattern, so I would appreciate any ideas for solving these questions.

edit:
I have attempted questions 2 and 3 by considering the opposite case, as suggested by one of the comments. But I got a counter intuitive result. Would appreciate any help in verifying if my method is correct.

2) $P($Rolling 12 dice and getting at most 1 '3'$)
= (5/6)^{12} + (\binom{12}{1})(1/6)(5/6)^{11}$

$P($Rolling 12 dice and getting at least 2 '3's$)
= 1 - ($ANS ABOVE$) = 0.6187$

3) $P($Rolling 18 dice and getting at most 2 '3's)

$= (5/6)^{18} + (\binom{18}{1})(1/6)(5/6)^{17} + (\binom{18}{2})(1/6)^2(5/6)^{16}$

$P($Rolling 18 dice and getting at least 3 '3's$)
= 1 - ($ANS ABOVE$) = 0.5973$

My answer for question 1 is $0.5981$, so this trend seems a bit counter-intuitive.