Finding probability -picking at least one red, one blue and one green ball from an urn when six balls are selected Six balls are to be randomly chosen from an urn containing $8$ red, $10$ green, and 12 blue balls.
What is the probability at least one red ball, one blue and one green ball is chosen?
Sample space = $\binom{30}{6}$
P = 1 - P(All red + All Green + All Blue + Only red and Green + Only Red and Blue + Only Green and Blue )
$$P = \Large 1 - \frac{\binom{8}{6} + \binom{10}{6} + \binom{12}{6} + \binom{18}{6} + \binom{22}{6} + \binom{20}{6}}{\binom{30}{6}}$$
According to this, I got $\large 1 - \frac{133099}{593775}$, 
which is $0.7758$?
Is my approach correct?

Your approach is not correct.
In 'Only red and green" you don't exclude situations "Only red" and "only green".
Also each of the situations "Only x" is counted twice - in situation "Only x and y" and "only x and z"
Thus the number of situations where there is a color missing should be:
$$\left(\binom{8}{6} + \binom{10}{6} + \binom{12}{6} + \binom{18}{6} + \binom{22}{6} + \binom{20}{6}\right)-2\left( \binom{8}{6} + \binom{10}{6} + \binom{12}{6} \right)=$$
$$= \binom{18}{6} + \binom{22}{6} + \binom{20}{6}-\left( \binom{8}{6} + \binom{10}{6} + \binom{12}{6} \right)$$
