Suppose set $S$ is composed of three types of elements. Let the number of these four types of elements in set $S$ be $S_a$, $S_b$, and $S_c$, so $|S| = S_a + S_b + S_c$.
Suppose set $T$ is a random subset of $S$ with size $t$. By random, it means each element of $S$ has the same chance to be selected in $T$. Then, denote the number of these three types of elements in set $T$ as $T_a$, $T_b$, and $T_c$.
Now does $\operatorname{E}(T_a/(T_a + T_b))$ equal to $S_a/(S_a + S_b)$? Or in another way, is $T_a/(T_a + T_b)$ an unbiased estimator of $S_a/(S_a + S_b)$?
Update:
I proved it numerically by calculating
$$ \sum \dfrac{T_a}{T_a + T_b}\dfrac{\binom{S_a}{T_a}\binom{S_b}{T_b}\binom{S_c}{t - T_a - T_b}}{\binom{S_a + S_b + S_c}{t}} = \dfrac{S_a}{S_a + S_b} $$
with a computer. But is there an elegant mathematical way to prove this?
Here's an answer from my friend.
$$ \begin{split} \operatorname{E}\left(\dfrac{T_a}{T_a + T_b}\right) &= \sum_{T_c = 0}^t \dfrac{1}{T_a + T_b}\dfrac{\binom{S_c}{T_c}}{\binom{|S|}{t}}\sum_{T_a + T_b = 0}^{t-T_c}T_a\binom{S_a}{T_a}\binom{S_b}{T_b} \\ &= \sum_{T_c = 0}^t \dfrac{1}{T_a + T_b}\dfrac{\binom{S_c}{T_c}}{\binom{|S|}{t}}\sum_{T_a + T_b = 0}^{t-T_c}S_a\binom{S_a - 1}{T_a - 1}\binom{S_b}{T_b} \\ &= S_a\sum_{T_c = 0}^t \dfrac{\binom{S_c}{T_c}}{\binom{|S|}{t}}\dfrac{1}{t - T_c}\binom{S_a + S_b - 1}{t - T_c - 1} \\ &= S_a\sum_{T_c = 0}^t \dfrac{\binom{S_c}{T_c}}{\binom{|S|}{t}}\dfrac{1}{S_a + S_b}\binom{S_a + S_b}{t - T_c} \\ &= \dfrac{S_a}{S_a + S_b}\dfrac{1}{\binom{|S|}{t}}\sum_{T_c=0}^t\binom{S_c}{T_c}\binom{S_a + S_b}{t-T_c}\\ &= \dfrac{S_a}{S_a + S_b}\dfrac{\binom{|S|}{t}}{\binom{|S|}{t}} \\ &= \dfrac{S_a}{S_a + S_b} \end{split} $$