= Solution
Use the intended model of uniform draws and a replacement decision independent of the first ball's colour, with <probability> $r$. The marginal specification $P(R)=r$ alone would not determine the experiment if replacement depended on that colour. This independence assumption is used throughout the question.
Write $\overline B_j$ for a non-blue draw and $\overline R$ for no replacement. The <sample space> consists of the eight ordered outcomes obtained by choosing one of $B_1,\overline B_1$, then one of $R,\overline R$, then one of $B_2,\overline B_2$. The <chain rule for probabilities> gives their <probabilities>:
$$
\boxed{\begin{array}{c|c}
\text{outcome}&\text{probability}\\\hline
B_1\cap R\cap B_2&4r/25\\
B_1\cap R\cap\overline B_2&6r/25\\
\overline B_1\cap R\cap B_2&6r/25\\
\overline B_1\cap R\cap\overline B_2&9r/25\\
B_1\cap\overline R\cap B_2&(1-r)/10\\
B_1\cap\overline R\cap\overline B_2&3(1-r)/10\\
\overline B_1\cap\overline R\cap B_2&3(1-r)/10\\
\overline B_1\cap\overline R\cap\overline B_2&3(1-r)/10
\end{array}}
$$
For example, the first entry is $(2/5)r(2/5)$. The fifth is $(2/5)(1-r)(1/4)$ because one blue ball has been removed. With no replacement after a non-blue first draw, the two remaining colours each have two balls, explaining the last two entries. The first four entries sum to $r$ and the last four to $1-r$, so the full <sample space> has total <probability> one. At $r=0$ or $1$, the corresponding impossible outcomes simply have zero <probability>.
Back to article page
