This is the oft-encountered problem of estimating a "confidence interval for a Binomial proportion". There are several different techniques, which vary in their assumptions and are well described on the Wikipedia page.
For all reasonable approaches, the precision (i.e. width of the confidence interval) will depend on
- your sample success ratio (i.e. $\hat{p}=\frac{422}{n}$)
- the number of measurements (i.e. $n=422+584$)
- the desired "confidence level" (i.e. a number $\alpha\in(0,1)$ giving the fractional coverage of the associated interval estimate)
(Note: I put some terms in quotes to try and steer clear of the thorny issues surrounding confidence-intervals vs. credible-intervals, $p$-values, etc.)
Update: At the request of the OP, here is some clarification on how the interval estimate would relate to the numerical precision.
For simplicity, I will assume that "$d$ significant digits" can be interpreted as "$\pm\,5\times 10^{-(d+1)}$", i.e.
$$
\hat{p} = 0.722 \implies p = \hat{p} \pm 0.0005 \implies 0.7215 < p < 0.7225
$$
where $\hat{p}$ is the estimate of the true value $p$. So a significance of $d$ digits roughly corresponds to an interval width of $10^{-d}$.
Again for simplicity, assume we use the "normal approximation", so for a given "confidence level" (specified as a "$z$-score") we have interval width
$$
\frac{\Delta{p}}{z}=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}
$$
So, for example if we use a coverage probability of 95%, corresponding to $z=1.96$ (roughly 2 standard deviations), then for $\hat{p}=\frac{422}{n}$ and $n=422+584$, we have
$$
\Delta{p}\approx 3 \times 10^{-2} = \mathrm{O}[10^{-2}]
$$
so, keeping things simple, we might say that there are roughly two significant digits. So while nominally we have $\hat{p}\approx 0.419483101391650$, from these calculations $p\approx 0.42$ would be a reasonable inference (with the $\pm 0.03/2$ implicit in the significant-digit truncation).
