Expected number of balls remaining A box contains $w$ white balls and $b$ black balls. Each time we pick a ball without replacement until there are no white balls left. What is the expected number of black balls remaining in the box ?

bof's answer is correct. I just want to add another solution here for fun.
Denote $B$ as the number of remaining black balls. Consider the picking as a permutation of all the balls.If there are $k$ black balls remaining, where $0 \le k \le b$, we know that the $(w + b - k)$th ball is white. Among the first $w+b-k -1$ balls, there are $w - 1$ white balls and $b - k$ black balls. So,
$P(B=k) = \frac{\binom{w+b-k-1}{b-k}w!b!}{(w+b)!}$.
Therefore,
$E(B) = \sum_{k=0}^{b}kP(B=k) = \frac{w!b!}{(w+b)!}\sum_{k=0}^{b}k\binom{w+b-k-1}{b-k}=\frac{b}{w+1}$.
