Solution
= Solution
There are two equally likely balls bearing label eight, one of each colour. The <conditional probability> that the removed ball was green is therefore $1/2$. By the <law of total probability>,
$$
\mathbb P(\text{second green}\mid X_1=8)
=\frac12\frac9{19}+\frac12\frac{10}{19}
=\boxed{\frac12}.
$$
Both possible first colours must be included because the observed label alone does not determine the colour.