= Solution
Every urn contains $n-1$ balls. Let $k=n-r$ be its number of blue balls; choosing an urn uniformly makes $k$ uniform on $0,\ldots,n-1$. Conditional on that urn, the <probability> of a blue first draw is $k/(n-1)$, so the <law of total probability> gives
$$
\boxed{\mathbb P(B_1)=\frac1n\sum_{k=0}^{n-1}\frac{k}{n-1}=\frac12.}
$$
Drawing twice without replacement requires $n\ge3$. Given $k$, the <probability> that both draws are blue is $k(k-1)/((n-1)(n-2))$, and hence
$$
\mathbb P(B_1\cap B_2)=\frac{\sum_{k=0}^{n-1}k(k-1)}{n(n-1)(n-2)}=\frac13.
$$
Taking the ratio defining <conditional probability> gives \b[the second requested probability]
$$
\boxed{\mathbb P(B_2\mid B_1)=\frac{1/3}{1/2}=\frac23.}
$$
Although a blue first draw removes a blue ball from its urn, it also makes an urn rich in blue balls more likely. Averaging with this changed conditional distribution is essential.
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