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)

In essence you are asking for the distribution of the variable Z = X/Y, where Z can take values in $\left\{ 0, \frac{1}{m}, \ldots, \frac{n}{1}, \infty \right\}$, where the fractions are of the form $\frac{k}{l}$ with $0 \leq k \leq n$, and $1 \leq l \leq m$, and I'm assuming the convention $\frac{k}{0} = \infty$, which as Guy pointed out means this will be undefined, and the standard deviation will be infinite. So instead condition that this cannot happen
$$
\mathbb{P}\left[Z = k/l | Y \neq 0 \right] = \frac{\mathbb{P}[Xl = YK \cap Y \neq 0]}{\mathbb{P}[Y \neq 0]}
$$
So then this gives,
$$
\begin{align}
\mathbb{P}\left[Z = k/l | Y \neq 0 \right] & = (1- (1-q)^m) \mathbb{P}[ Xl = Yk \cap Y \neq 0]\\
& = (1-(1-q)^m) \sum p^i(1-p)^{(n-i)}q^j(1-q)^{(m-j)}
\end{align}
$$
where the sum is over all pairs $(i,j)$ with $0 \leq i \leq n$, $0 < j \leq m$ such that $i/j = k/l$... All in all it looks like its going to be a messy task to compute this sum, and to find the standard deviation. 
