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
 A: 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 $p\approx 0.13$,
A: Split the event into two disjoint events:

*

*$E_1$: When you get atleast $3G$ and exactly $0R$,

*$E_2$: When you get atleast $3G$ and exactly $1R$.

$P(E_1)=\sum_{k=3}^6\binom 6k(1/4)^k(2/4)^{6-k}$ since you pick $k$ out of $6$ tries to get $k\ge3$ green balls and on the remaining $6-k$ tries, you get either orange or yellow ball.
$P(E_2)=\sum_{k=3}^5\binom 6k(1/4)^k\binom{6-k}1(1/4)(2/4)^{6-k-1}$ since you pick $k$ out of $6$ tries to get $k\ge3$ green balls, $1$ try out of remaining $6-k$ tries to get a red ball and on the remaining $6-k-1$ tries, you get either orange or yellow.
The required probability is $P=P(E_1)+P(E_2)$ which indeed turns out to be $539/4096$.
A: $$1Red(3 Green+4Green+5Green)+\\
0Red(3Green+4Green+5Green+6Green)$$
For example, 1Red gives a factor of ${6\choose1}=6$.  For that case, 3Green has a count of ${5\choose3}2^2=40$ because there are two Other balls.
