# Probability of Event - 3 persons in a club win a lottery and also happen to have all black cars

In a club every year they celebrate a special anniversary with a lottery.

The Club has 3,000 members.

3 Club members win in the lottery every year (3 different prizes, but that is not material here).

In that Club, 100 members happen to have black cars.

Obviously the color of the car of a member and whether a member will win in the lottery are independent events.

The lottery is made with replacement. I.e., one member can win all 3 prizes if one is very lucky.

This year all three winners of the lottery happened to have also black cars.

What was the probability of this event?

Would it be reasonable to say that it was $$((100/3000) * (1/3000))^3?$$ Explain.

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.