= Solution
Distinguish the $N$ balls by labels. For a fixed <position> $n$, each label appears there in exactly $(N-1)!$ of the $N!$ equally likely <permutations>. Since $N_B$ labels are black,
$$
\boxed{\mathbb P(B_n)=\frac{N_B(N-1)!}{N!}=\frac{N_B}{N},
\qquad 1\leq n\leq N.}
$$
This is the same marginal law at every <position>, even the last one. The argument averages over earlier draw outcomes; it does not assert that a <conditional probability> after a specified history remains $N_B/N$. The original PDF uses $B_n$, correcting the transcription's uppercase subscript $B_N$.
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