So you want to know the probability of getting three or more rolls of 4+ on nine dice?

This is easier stated as 1 minus the probability of only getting 0,1 or 2 such successes. The binomial theorem gives us
1 way to get 0 successes
9 ways to get only 1 success
9*8/2=36 ways to get only 2 successes
p=1-((1*(3/6)^9)+(9*(3/6)*(3/6)^8)+(36*(3/6)^2*(3/6)^7))
=1-(46*(1/2)^9)
that's a 91 percent chance of success if I've understood the question.