jmc

There are three different possibilities for our first decision, each corresponding to which container we choose. So, if we choose container A, with $\frac{1}{3}$ probability, we have a $\frac{6}{10}=\frac{3}{5}$ probability of drawing green, which means we have a $\frac{1}{3}\cdot \frac{3}{5}=\frac{1}{5}$ of picking Container A and then picking a green ball. Similarly for container B the probability is $\frac{1}{3}\cdot \frac{4}{10}=\frac{2}{15}$, and the same for container C. So, the total probability is $\frac{1}{5}+\frac{2}{15}+\frac{2}{15}=\boxed{\frac{7}{15}}$.