at least one passenger in each station

Question

I have the following problem: There is a bus with $30$ passengers that are going to get off at the next $3$ stations $S_1,S_2,S_3$.Find the probability that at least one passenger gets off the bus at each of the stations.I tried to find the probability of the complement event but didn't manage to complete the calculation.Any ideas?

Answer 1

If its random (which seems stupid), then you can use Sterling numbers of the second kind to count all the ways in which the people could get off.

$S(30,1) = 1$

$S(30,2) = 536,870,911$

$S(30,3) = 34,314,651,811,530$

There are $\binom{3}{1}$ ways to get off at 1 stop, and $\binom{3}{2}$ ways to get off at 2 stops, so we have to multiply $S(30,1)$ and $S(30,2)$ by 3. Additionally There are $3!$ ways to order each of the outcomes of $S(30,3)$ and $2!$ to order $S(30,2)$.

So then we have the probability someone gets off at each station as

$\cfrac{3! \cdot S(30,3)}{3! S(30,3) + 2! \binom{3}{2 }S(30,2) + \binom{3}{1}S(30,1)}$

$\cfrac{205887910869180}{ 205887910869180 + 3221225466 + 3} = 999984354714862$