How to obtain the Standard Deviation of a ratio of independent binomial random variables? X and Y are 2 independent binomial random variables with parameters (n,p) and (m,q) respectively.
(trials, probability parameter)

There probably isn't a closed-form formula for this.
But $X$ has mean $np$ and standard deviation $\sqrt{np(1-p)}$, and $Y$ has mean $mq$ and standard deviation $mq(1-q)$.
Now you need a simple fact: if $X$ has mean $\mu$ and standard deviation $\sigma$, then $\log X$ has mean approximately $\log \mu$ and standard deviation approximately $\sigma/\mu$.  This can be derived by Taylor expansion. Intuitively, $X$ "usually" falls in $[\mu-\sigma, \mu+\sigma]$ and so $\log X$ "usually" falls in $[\log (\mu-\sigma), \log (\mu+\sigma)]$. But we have
$$ \log (\mu \pm \sigma) = \log \Big(\mu(1 \pm \sigma/\mu)\Big) = \log \mu \pm \log(1 \pm \sigma/\mu) \approx \log \mu \pm \sigma/\mu $$
where the approximation is the first-order Taylor expansion of $\log (1+x)$ for $x$ close to zero.
Therefore $\log X$ has mean approximately $\log np$ and standard deviation approximately $\sqrt{(1-p)/np}$. Note that for the Taylor expansion above to be sufficient, $\sigma/\mu=\sqrt{(1-p)/np}$ must be close to zero. Similarly $\log Y$ has mean approximately $\log mq$ and standard deviation approximately $\sqrt{(1-q)/mq}$.
So $\log X - \log Y = \log X/Y$ has mean approximately $\log(np/mq)$ and standard deviation approximately
$$ \sqrt{{(1-p) \over np} + {(1-q) \over mq}}. $$
But you asked about $X/Y$. Inverting the earlier fact, if $Z$ has mean $\mu$ and standard deviation $\sigma$, then $e^Z$ has mean approximately $e^{\mu}$ and standard deviation approximately $\sigma e^\mu$. Therefore $X/Y$ has mean approximately $np/mq$ (not surprising!) and standard deviation approximately
$$ \left( \sqrt{{(1-p) \over np} + {(1-q) \over mq}} \right)  {np \over mq}. $$
This approximation works well if $p,q$ and/or $m,n$ are not too small (see Taylor expansion explanation in the middle of this answer).
