Suppose I have a source of red and blue coins that have a red composition of absolute percentage $a_r$ with uncertainty $\sigma_r$, and similar measures for blue. If I take $N$ coins randomly from the source, it's trivial to expect I have (approximately) $N a_r$ red coins and $N a_b$ blue coins. To what uncertainty though? That is, how well to I expect to have these numbers?
I would blindly assume that I would have an uncertainty of $\sqrt{a_r(1-a_r)N}/\sqrt{N}=\sqrt{a_r(1-a_r)}$ (standard error from standard deviation) as I'm drawing from a binomial distribution (either got the color of interest or not). I would assume $\sqrt{a_b(1-a_b)}$ for the blues as well. I can't seem to convince myself though this is correct. Also, this pays no heed to the uncertainty in the percentages! Where would these come into play? What if I had $n$ different colors/percentages/uncertainties?
For the sake of application, lets simply assume the following example values:
\begin{align*} \text{Red:} \quad a_r&=0.5 \quad \sigma_r=0.09 \\ \text{Blue:} \quad a_r&=0.4 \quad \sigma_r=0.04 \\ \text{Green:}\quad a_r&=0.1 \quad \sigma_r=0.01 \end{align*} $$ N=100 $$
Alright, here's my attempt in general detail. Correct me if I'm wrong! Suppose we have $n$ different variations in our source with relative compositions (and their associated uncertainties) of $a_i \pm \sigma_i$. We then draw $N$ pieces from our large source and assume we have $N_i$ objects of the $i$th variety to an extended uncertainty of $\sigma_{Ni}^*$: \begin{align*} N_i &= E(i) \\ &= N a_i \end{align*} \begin{align*} \sigma_{Ni} &= \frac{SD(i)}{\sqrt{N}} \\ &= \frac{\sqrt{a_i(1-a_i)N}}{\sqrt{N}} \\ &= \sqrt{a_i(1-a_i)} \\ \therefore\quad \sigma_{Ni}^* &= \sqrt{a_i(1-a_i) + \sigma_i^2} \end{align*} I've made no inclusion of rounding matters. For a 'realistic' $N_i$, then $\text{Round}($N_i$)$ should be taken independently. I say independently because the uncertainty here would not make sense to report with the altered quantity.