# Elevator probability calculation

I'm learning probability, specifically techniques of counting, and need help with the following problem :

There are $5$ people in an elevator, $4$ floors in the building and each person exits at random. Find the probability that :

$(1)$ no one exit on the first floor;

$(2)$ at least one person exit on the first floor and at least one person on the second floor.

Since I'm having difficulties for $(2)$, I'm going to share my work for $(1)$.

$(1)$ The number of ways to assign the $3$ remaining floors to the $5$ people is $3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 3^5$ because for each person we can choose one of the $3$ remaining floors. By the same argument, there are $4^5$ ways to assign $4$ floors to $5$ people. Therefore, the requested probability is $$\frac{3^5}{4^5}.$$

Is my work correct for $(1)$? Any help for $(2)$ will be greatly appreciated.

• The statement "each person exits at random" is kind of vague. At each floor, does each person have a certain probability of leaving, or does each person, at the beginning, pick a random floor to get off at? – Frpzzd Jun 16 '17 at 20:55
• @Nilknarf I'm translating this exercise from my notes which are in greek so this is possibly not the most accurate translation. I think what is meant here is that all floors are equally likely. – user347616 Jun 16 '17 at 21:04
• Okay, I'll leave you an answer. :) – Frpzzd Jun 16 '17 at 21:05
• Your work is correct on 1). – Doug M Jun 16 '17 at 21:14

Nobody gets off on the First floor: $243$

Nobody gets off on the Second floor: $243$

Nobody gets off on at either floor: $2^5 = 32$

Note that nobody gets off at either floor is a subset of both nobody gets of at the first floor it is also a subset of nobody gets off at the second floor.

Nobody gets off on the First floor or Nobody gets off on the second floor: $243 + 243 - 32 = 454$

We have to subtract 32 to avoid double counting.

Somebody gets off at both the first floor and the second floor $= 4^5 - 454 = 570$