Drawing balls with two features Suppose that I have a total of 1000 balls in a bag. 400 balls are labeled a and remaining 600 balls are labeled b. In addition, within the same 1000 balls, 60 balls are white and remaining 940 balls are black. 
Now, if I conduct an experiment of picking a ball one after other without replacement, lets say 30 times, how can I compute the probability that I would have at least 20 balls that are white and labeled a? I would like to basically know how probable is this event of picking 20 balls that are white and labeled a out of 30 events, given this setup, due to chance.
And its not a homework question.
Thank you!
 A: As people pointed out in the comments, your description doesn't say whether the number of balls which are white and labelled $a$ is $60$ (all white are labelled $a$), $24$ (the label is independent of the color), or $0$ (all white are not labelled $a$). However, what you can say is that it is quite unlikely to get $20$ out of $30$ white balls labelled $a$ if the balls are chosen uniformly. This would be most probable with $60$ white balls labelled $a$, and it's still astronomically unlikely.
The relevant distribution is a hypergeometric distribution. If you assume that there are $60$ white balls labelled $a$ in the population of size $1000$, then the chance to get exactly $20$ out of $30$ is 
$$\frac{{60 \choose 20}{940 \choose 10}}{1000 \choose 30} \approx 2.44 \times 10^{-19}. $$
However, instead of the exact chance to get $20$, you should consider the chance to get $20$ or more, which is $2.49 \times 10^{-19}$. This is an upper bound for the chance that $20$ out of $30$ will be both white and labelled with $a$.
