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)
 A: 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).
A: If $n$ and $m$ go to infinity while $p$ and $q$ are fixed, then the ratio $R=X/Y$ is well defined on an event of probability $1-o(1)$. 
On this event (or conditionally on this event, since the two asymptotically equivalent), Edgeworth expansions of $X$ and $Y$ show that the expectation of $R$ behaves like $\dfrac{np}{mq}$ and that its variance behaves like
$$
\left(\frac{np}{mq}\right)^2\left(\frac{1-p}{np}+\frac{1-q}{nq}\right).
$$
A: 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. 
A: This shoud be a comment over Michael Lugo's answer. The Taylor expansion approach can be done directly, without taking logarithm and inverting.
In general, for a diferentiable $g(y)$ and a small variance random variable $Y$ we have
$$ Z = g(Y) \approx g(\mu_Y) + g'(\mu)(Y-\mu_Y) + \frac{1}{2}g''(\mu)(Y-\mu_Y)^2 + \cdots$$
$$ \mu_Z \approx g(\mu_Y) + \frac{1}{2}g''(u_Y)\sigma_Y^2 + \cdots \tag 1 $$
and
$$ \sigma_Z^2 \approx [g'(\mu_Y)]^2   \sigma_Y^2  \tag 2$$
In particular, for $g(Y)=1/Y$:
$$  \mu_Z \approx \frac{1}{\mu_Y} +  \frac{\sigma_Y^2}{\mu_Y^3}  $$
$$  \sigma_Z^2 \approx \frac{\sigma_Y^2}{\mu_Y^4}$$
Then, assuming $X,Y$ are independent:
$$E[X/Y] =E[X] E[1/Y] \approx \frac{\mu_X}{\mu_Y} + \mu_X\frac{\sigma_Y^2}{\mu_Y^3} \approx \frac{\mu_X}{\mu_Y}$$
Also, using this:
$${\rm Var} (X/Y) \approx \sigma_X^2 \frac{\sigma_Y^2}{\mu_Y^4} + 
\sigma_X^2 (\frac{\mu_X}{\mu_Y})^2 + \frac{\sigma_Y^2}{\mu_Y^4} \mu_X^2 \approx 
\frac{\mu_X^2}{\mu_Y^2}\left(\frac{\sigma_X^2}{\mu_X^2} + \frac{\sigma_Y^2}{\mu_Y^2}\right)$$
