= Solution
Urn $r$ has $n-r$ blue balls among $n-1$ balls. Averaging over the uniformly chosen urn,
$$
\mathbb P(\text{first blue})
=\frac1n\sum_{r=1}^n\frac{n-r}{n-1}
=\boxed{\frac12}.
$$
The probability that both removed balls are blue is
$$
\begin{aligned}
\mathbb P(B_1\cap B_2)
&=\frac1n\sum_{r=1}^n
\frac{(n-r)(n-r-1)}{(n-1)(n-2)}\\
&=\frac13.
\end{aligned}
$$
Therefore
$$
\boxed{
\mathbb P(B_2\mid B_1)
=\frac{\mathbb P(B_1\cap B_2)}{\mathbb P(B_1)}
=\frac23}.
$$
The first blue draw makes urns with more blue balls more likely, which explains why the conditional probability exceeds one half.
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