Approximately $0.15$ ( $360 / 2401$ .) (Assume that the choices of the $5$ passengers are independent. Also assume that the probability that a passenger chooses a particular floor is the same for all $7$ floors.)

Step-by-step explanation:

If there is no requirement that no two passengers exit at the same floor, each of these $5$ passenger could choose from any one of the $7$ floors. There would be a total of $7 \times 7 \times 7 \times 7 \times 7 = 7^{5}$ unique ways for these $5\!$ passengers to exit the elevator.

Assume that no two passengers are allowed to exit at the same floor.

The first passenger could choose from any of the $7$ floors.

However, the second passenger would not be able to choose the same floor as the first passenger. Thus, the second passenger would have to choose from only $(7 - 1) = 6$ floors.

Likewise, the third passenger would have to choose from only $(7 - 2) = 5$ floors.

Thus, under the requirement that no two passenger could exit at the same floor, there would be only $(7 \times 6 \times 5 \times 4 \times 3)$ unique ways for these two passengers to exit the elevator.

By the assumption that the choices of the passengers are independent and uniform across the $7$ floors. Each of these $7^{5}$ combinations would be equally likely.