Can this probability be computed more efficiently? Given an urn filled with 4 black balls, 4 red balls, the remaining being white, the probability of seeing AT LEAST one red ball AND AT LEAST one black ball when taking out 7 balls can be computed as follows:
$\sum_{r=1}^4 \left(\sum_{b=1}^4 \frac{{4 \choose r}{4 \choose b}{52 \choose 7-r-b}}{{60 \choose 7}}\right)$
Now, someone recently commented saying that this could be computed much more efficiently, but I never got an answer as to how. Can somebody tell me how else this probability can be computed, which would be more efficient as well?
 A: Hint:
Let $A$ be the event of seeing at least one red ball, and $B$ the event of seeing at least one black ball.
$$P(A \cap B) = 1-P((A \cap B)^c) = 1-P(A^c \cup B^c) = 1-[P(A^c) + P(B^c) - P(A^c \cap B^c)].$$
$P(A^c)$ is the probability of seeing no red balls, and $P(B^c)$ is the probability of seeing no black balls.
A: More than a hint
Angryavian above shows us the general principle, "Complementary probabilities". Instead of explicitly calculating the probability for an event we can calculate the probability for anything else happening and use the fact that all probability will add up to 1.
So for the complementary event we get 
$$1 - \sum_{r=0}^1 \left(\sum_{b=0}^1 \frac{{4 \choose r}{4 \choose b}{52 \choose 7-r-b}}{{60 \choose 7}}\right) + \left(\frac{{4 \choose 1}{4 \choose 1}{52 \choose 7-1-1}}{{60 \choose 7}}\right)$$
This will be like having to calculate 3 terms instead of... $4\cdot 4 = 16$. Also please note that the (1,1) is first subtracted and then added. So what we actually want to calculate is $$1-\sum_{r=0,b=0}^{r+b\le 1}\left(\frac{{4 \choose 1}{4 \choose b}{52 \choose 7-r-b}}{{60 \choose 7}}\right)$$
