Solution
= Solution
Let $X$ count green balls. Sampling without replacement gives a <hypergeometric distribution>. The possible values are $0,1,2$, and counting equally likely three-element samples gives
$$
P(X=k)=\frac{\binom2k\binom3{3-k}}{\binom53}.
$$
Therefore
$$
\boxed{P(X=0)=\frac1{10},\qquad P(X=1)=\frac35,\qquad P(X=2)=\frac3{10}}.
$$
Their sum is $(1+6+3)/10=1$, as required for a <probability distribution>.