This seems very basic, but I can't find a clear statement of it.
Suppose I have a population of $N$ balls which are red, white, and blue in some proportion. If I take a sample of $S$ balls ($S \ll N$) and use $S_r/S$, $S_w/S$ and $S_b/S$ to estimate the fraction of each colour in the original population, what can be said about the variance of $S_r/S$?
On
Let $p_r$ be the proportion of red balls out of $N.$ Filling in the details of the previous answer, the variance of a binomial is:
$$Sp_r(1-p_r)$$
And for sampling with replacement, the variance of the hypergeometric is
$$Sp_r(1-p_r)\frac{N-S}{N-1}$$
The last term (the finite population correction factor) will be close to $1$ if $S<<N.$
With the assumption that $S << N$ we can reasonably treat this as sampling with replacement, which makes the math a lot easier.
We want to draw red balls. We can model each draw as an iid draw from a Bernoulli distribution with probability of success $p$, i.e. our draws $X_1, X_2, ..., X_n \sim \ \ iid \ \ Bern(p)$. Note how assuming sampling with replacement lets us assume independence.
For these $n$ draws, let $S_r$ be the total number of red balls observed. Then $S_r = \sum_{i=1}^{n} X_i$ has a well-known distribution. From this we can get the variance of $S_r$ which we can then divide by $n$ in my notation or $S$ in yours [making sure to properly follow the variance rules].