Green and Blue?

The probability of picking out the two balls known to be blue is $\frac{1}{6}$ , in which case, the probability of getting a green and a blue is zero.

The probability of picking one of the known blue balls and a "random ball" is $\frac{2}{3}$ , in which case the probability that it is a green and a blue is $\frac{1}{3}$ .

The probability of picking the two "random" balls is $\frac{1}{6}$ . In which case, the probability of picking a blue and a green is $\frac{2}{9}$ since there are two ways out of nine possibilities that one is blue and one is green.

Therefore, the probability of picking a green and a blue is:

P = $\frac{1}{6} \cdot 0 +\frac{2}{3} \cdot \frac{1}{3} + \frac{1}{6} \cdot \frac{2}{9}= \frac{7}{27}$

$7+27 = \boxed{34}$

The probability of the first ball being green is $\frac{1}{2} \times \frac{1}{3}$ = $\frac{1}{6}$ . Then the probability for the second ball being blue is $\frac{2}{3}$ + $\frac{1}{3} \times \frac{1}{3}$ = $\frac{7}{9}$ So the total probability of first picking a green ball and then a blue ball is $\frac{1}{6} \times \frac{7}{9}$ = $\frac{7}{54}$ and since the order of picking the balls does not matter we can multiply this by 2 (the count of variations of picking a green and a blue ball) getting the result $\frac{7}{27}$ . Therefor the answer is 7 + 27 = $\boxed{34}$