Complex probability problem

[aha]

I am stucked at this exercise for more than 6 hours. Could you solve this please:

There is 8 stages at the house. 4 people get in to the elevator at the ground floor. Compute the probabilty, that one person will exit the elevator at floor 2 and that the other three people will exit the elevator at the same floor (but not at floor 2)?



Thank you very much!

 

[tkhunny]

Small wonder you are struggling. There is not enough information.

Is there a distribution of some sort suggesting expected values for abandoning the elevator on each floor?

 

[pka]

There is 8 stages at the house. 4 people get in to the elevator at the ground floor. Compute the probabilty, that one person will exit the elevator at floor 2 and that the other three people will exit the elevator at the same floor (but not at floor 2)?

Let's agree that there are eight floors at which the people can get off. We do not care who gets off just that some one gets off at two and the rest get at the same other floor. There \(\dbinom{4+8-1}{4}=\dfrac{11!}{4!\cdot 7!}=330\) way those four people can exit with no restrictions whatsoever.
Of those there are ten ways for one person to get off on floor two and the other three to all exit on another floor.