Method 1
The probability of R, G + Probability of R, B + Probability of B, G
Above equation covers All red, All blue and All Green twice so we subtract it once, Thus we get:
The probability of R, G + Probability of R, B + Probability of B, G - ALL R - ALL G - ALL B
1-$\frac{C(18,6)+C(20,6)+C(22,6)-C(8,6)-C(10,6)-C(12,6)}{C(30,6)}$=0.7797
1- ( all red + all green + all blue + red,green + red,blue+ green,blue) this is incorrect. It has several repetition
Method 2
let <R, G, B> be a triplet
now we first take One R, One G, One B now we have 3 balls left to pick
<3,0,0>, <2,0,1>, <2,1,0>,<1,2,0>,<1,0,2>,<1,1,1>,<0,2,1>,<0,1,2>,<0,3,0>,<0,0,3>
So we have 10 triplets which are same if we solve generating function
Xr+Xg+Xb=6
($x+x^{2}+...+x^{8}$)*($x+x^{2}+...+x^{10}$)*($x+x^{2}+...+x^{12}$)=6
($1+x^{1}+...+x^{7}$)*($1+x^{1}+...+x^{9}$)($1+x^{1}+...+x^{11}$)=3
$\left ( \frac{1-x^{8}}{1-x} \right )$*$\left ( \frac{1-x^{10}}{1-x} \right )$*$\left ( \frac{1-x^{12}}{1-x} \right )$=3
$\left ( 1-x^{8} \right )$*$\left ( 1-x^{10} \right )$*$\left ( 1-x^{12} \right )$*$\left ( 1-x \right )^{-3}$=3
$\rightarrow$C(5,3)=10
Initially, we took 1 R, 1 G, 1 B adding it to triplet we get
<4,1,1>,<3,1,2>,<3,2,1>,<2,3,1>,<2,1,3>,<2,2,2>,<1,3,2>,<1,2,3>,<1,4,1>,<1,1,4>
so total favorable case is 4,1,1=C(8,4)*C(10,1)*C(12,1)=8400
3,1,2=36960
3,2,1=30240
2,3,1=40320
2,1,3=61600
2,2,2=83160
1,3,2=63360
1,2,3=79200
1,4,1=20160
1,1,4=39600
=463000 combination in which we have at least 1 R, 1 G, 1 B
Probability=$\frac{463000}{C(30,6)}$=0.7797