Hypergeometric Distribution and Combinations Suppose there are 30 balls in an urn, 2 red, 2 black and the other 26 white. 
After selecting at random 4 balls without replacing, what is the probability of having at least one red and at least one black ball? Also, what if I were to continue picking balls one by one after the initial 4, how would I go about calculating that. 
I know how to use the hypergeometric distribution to know the probability of getting at least one red ball, but I don't know how to compute for one red and one black. 

I would say (analogous to hypergeometric probability distribution): 
When four draft:
$\displaystyle P(red = r, black = b)=\frac{{2 \choose r}{2 \choose b}{26 \choose 4-r-b}}{{30 \choose 4}},\quad r,b\in\langle 0,2 \rangle$
$\displaystyle P(red >0, black >0)=\sum_{r=1}^2 \left(\sum_{b=1}^2 \frac{{2 \choose r}{2 \choose b}{26 \choose 4-r-b}}{{30 \choose 4}}\right)=\frac{281}{5481}$
When 'n > 4' draft
$\displaystyle P(red >0, black >0)=\sum_{r=1}^2 \left(\sum_{b=1}^2 \frac{{2 \choose r}{2 \choose b}{26 \choose n-r-b}}{{30 \choose n}}\right)$
E.g.:  n = 28, P = 0.995; n = 29, P = 1; n = 30, P = 1
