Solution
= Solution
Let $R_1,R_2$ be the <events> that the first and second birds are ringed. Under <sampling without replacement>,
$$
\mathbb P(R_1\cap R_2)=\frac4{10}\frac39=\frac2{15}.
$$
Thus the <conditional probability> is \b[$\displaystyle\mathbb P(R_1\mid R_2)=\frac{2/15}{2/5}=\frac13$]. Alternatively, once a particular ringed second bird is specified, three of the nine possible first birds are ringed.