Also, we can assume everyone's probability of exiting at a given floor is the same.
Here's my work: let's pick two people to exit a floor alone. There are $ 6 \choose 2$ ways to do this. The first person can exit at $10$ different floors and the second one on any of the remaining $9$. Now, the remaining four people can't exit any floor alone, and this can only happen if either they all exit together on a given floor, and there are $8$ possible floors possible to do this, or $2$ of them get off at one floor and $2$ get off at another (it can't happen that $3$ get off at one floor and $1$ at another because then $3$ people would have exited alone and not $2$): there are $4 \choose 2$ ways to pick the first pair, $8$ possible floors they can exit on and for the second pair there are then $7$ possible floors for them to exit. Since there are $10^6$ possible ways for them to exit the floors, the probability that only two of them exit a floor alone is given by:
$${6 \choose 2} \cdot \frac{10\cdot 9\cdot \left(8 + {4 \choose 2} \cdot 8 \cdot 7 \right)}{10^6}$$
Can someone check my work? I'd like to know the correct answer and if there are any flaws in my reasoning.
Best Answer
There is a slight mistake, happening in the sentence "there are ${4\choose2}$ ways to pick the first pair".
Instead you should say: The oldest of the remaining $4$ passengers has three ways to choose his mate, and then these two can descend at $8$ floors, whereas the last two people have $7$ floors to choose from.
In this way the correct answer is $$\frac{{6 \choose 2} \cdot 10\cdot 9\cdot \left(8 + 3 \cdot 8 \cdot 7 \right)}{10^6}=0.2376\ .$$