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Answer:
Approximately [tex]0.15[/tex] ([tex]360 / 2401[/tex].) (Assume that the choices of the [tex]5[/tex] passengers are independent. Also assume that the probability that a passenger chooses a particular floor is the same for all [tex]7[/tex] floors.)
Step-by-step explanation:
If there is no requirement that no two passengers exit at the same floor, each of these [tex]5[/tex] passenger could choose from any one of the [tex]7[/tex] floors. There would be a total of [tex]7 \times 7 \times 7 \times 7 \times 7 = 7^{5}[/tex] unique ways for these [tex]5\![/tex] 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 [tex]7[/tex] 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 [tex](7 - 1) = 6[/tex] floors.
Likewise, the third passenger would have to choose from only [tex](7 - 2) = 5[/tex] floors.
Thus, under the requirement that no two passenger could exit at the same floor, there would be only [tex](7 \times 6 \times 5 \times 4 \times 3)[/tex] 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 [tex]7[/tex] floors. Each of these [tex]7^{5}[/tex] combinations would be equally likely.
Thus, the probability that the chosen combination satisfies the requirements (no two passengers exit at the same floor) would be:
[tex]\begin{aligned}\frac{(7 \times 6 \times 5 \times 4 \times 3)}{7^{5}} \approx 0.15\end{aligned}[/tex].