Probability of drawing at least 1 red, 1 blue, 1 green, 1 white, 1 black, and 1 grey when drawing 8 balls from a pool of 30? Given a pool of 30 balls (5 of each color). When drawing 8 balls without replacement, what is the probability of getting at least one of each color?
Related: Probability of drawing at least one red and at least one green ball.
When drawing more than 2 colors you need to exclude overlapping 'hands'. Thus when finding the probability of drawing no red, you can have a hand made up of blue, green, white, black and grey. But when you are determining the probability of drawing no blue you draw from red, green, white, black, grey. So you need to exclude all green, white, black, grey hands as they have already been counted. And the same for the other colors as well.
The other complexity of the problem is that since there are only 5 of each color, no draw will only include balls of the same color.

For the new version of the problem, we can use Inclusion-Exclusion:
If $E_i$ is the set of draws without at least one ball of color i, we have
$|E_1\cup\cdots\cup E_6|=\sum|E_i|-\sum|E_i\cap E_j|+\sum|E_i\cap E_j\cap E_k|-\cdots$
$=\binom{6}{1}\binom{25}{8}-\binom{6}{2}\binom{20}{8}+\binom{6}{3}\binom{15}{8}-\binom{6}{4}\binom{10}{8}$,
so the probability will be $\displaystyle1-\frac{\binom{6}{1}\binom{25}{8}-\binom{6}{2}\binom{20}{8}+\binom{6}{3}\binom{15}{8}-\binom{6}{4}\binom{10}{8}}{\binom{30}{8}}=\frac{5,000}{26,013}$.
