You have to account for each car and person. It would be the product of the probability for three people having black cars (assuming each member owns a different car),
$$
1. P(3\text{ black cars})=\frac{100}{3000}\frac{99}{2999}\frac{98}{2998}=\frac{100!}{3000!}\frac{(3000-3)!}{(100-3)!},
$$
with the probability of each winning the lottery, three times, but it can't be the same person.
$$
2. P(3 \text{ distinct winners}) = \frac{1}{3000}\left(\frac{1}{3000}\frac{2999}{3000}\right)\left(\frac{1}{3000}\frac{2998}{3000}\right).
$$
Explaining:
$1$. The probability of the first winner to have a black car is simply the number of black cars divided by the total number of members, as your intuition says. But then that person and that car are no longer an option, so there remains $99$ black cars and $2999$ members to choose, and so on.
$2$. Since we allow for repeated wins but we want three distinct winners, in each lottery draw we must require that the previous winners do not win, which, for the second draw, is the probability of a winner times the probability that this wasn't the previous winner, so $\frac{1}{3000}\frac{(3000-1)}{3000}$. For the third draw, we require that the winner is not any of the two previous winners, so $\frac{1}{3000}\frac{(3000-2)}{3000}$.
In the end, if you multiply everything, you find that the probability for all three (distinct) winners owning black cars is
$$
\frac{100*99*98}{3000^6}<\left(\frac{100}{3000^2}\right)^3,
$$
slightly less than what you said.
