Probability of picking at least $3$ green balls and atmost $1$ red ball from $4$ different balls in $6$ tries 
I have four balls in the bag {, , ,}. What's the probability of picking (with replacement) $3$ or more green balls and $1$ or less red ball out of $6$ tries?

When I am running experiment programmatically, result is $0.13$. $4096$ all possible combinations and $539$ combinations with $3$ or more greens and $1$ or less red. $539/4096\approx0.13$.
Now I am wondering what would be a math/formulaic way to derive same result?
I have tried multiplication rule but $P(\text{green}\ge3)\times P(\text{red}\le1) = 0.09$ which makes no sense :(
Any help would be much appreciated

Note that out of 6 tries we have that
$$P(\text{green$=3$ and red$=1$})=\frac{6!}{3!1!2!}\left(\frac{1}{4}\right)^3\cdot \left(\frac{1}{4}\right)^1\cdot \left(\frac{2}{4}\right)^2.$$
Your probability is equal to
$$p=\sum_{j=3}^6 P(\text{green$=j$ and red$=0$})+\sum_{j=3}^5 P(\text{green$=j$ and red$=1$}).$$
Can you take it from here and compute the exact value of $p$?
P.S. I confirm that $p\approx 0.13$,
