I need help with all problems pls. i don't know for the first one.
1 answer:
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Answer:
Approximately (
.) (Assume that the choices of the
passengers are independent. Also assume that the probability that a passenger chooses a particular floor is the same for all
floors.)
Step-by-step explanation:
If there is no requirement that no two passengers exit at the same floor, each of these passenger could choose from any one of the
floors. There would be a total of
unique ways for these
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 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 floors.
Likewise, the third passenger would have to choose from only floors.
Thus, under the requirement that no two passenger could exit at the same floor, there would be only 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 floors. Each of these
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:
.