How many balls should I pull out to be sure in resulted distribution with 99% probability? I have a bag with infinite count black and white balls. I don't know the distribution between black and white balls.
How many balls should I pull out to be sure in resulted distribution with 99% probability?
F.e. I'm pulling out 1000 balls. There are 300 white and 700 black ones. Are those 1000 balls enough to say: there are 30% white and 70% black balls in this infinite bag with 99% probability (or maybe accuracy)?
 A: For the way you've stated the problem, you would calculate the error E you could tolerate at the $99\%$ level of confidence.
$$E = z\cdot \sqrt{\frac{p(1-p)}{n}}$$
$$E = 2.58\cdot \sqrt{\frac{0.7\cdot 0.3}{1000}}$$
$$E = .03739$$
So for a proportion of $0.7$ the $99\%$ confidence interval would be $(0.66261, 0.73739)$, so for $n = 1000$, expect the number of black to be in the interval $(663, 737)$.  
I should add an interpretation of what this result means. Given a true proportion of $0.7/0.3$ of black/white balls, $99$ out of $100$ samples of $1000$ would contain between $663$ and $737$ black balls. Merry Xmas.
A: In connection with the given answer by @Phil H, you can use Cochran's formula to assess in a reasonable and simple way the minimum sample size needed (N) to determine the parameters of the distribution, given a level of precision and some parameter estimates:
$$N=\frac{Z^2\,p(1−p)}{\epsilon^2},$$ 
where $\epsilon$ is the desired level of precision (0.01 in your case), Z is the Z-score that corresponds to the desired confidence interval (with a 99% confidence interval, $Z=2.58$), p is an estimate of the proportion of the property being observed (0.3 in your example). By using such formula you will get that $$N=13978,$$ approximately. 
