Solution
= Solution
Let $Y$ indicate whether the draw from the second bag is green. It is independent of $X$, with $P(Y=1)=1/3$ and $P(Y=0)=2/3$. Exactly one green overall arises from the disjoint alternatives $(X,Y)=(1,0)$ or $(0,1)$. Therefore
$$
\boxed{P(X+Y=1)=\frac35\frac23+\frac1{10}\frac13=\frac{13}{30}}.
$$