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.

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 $= 4^5 - 454 = 570$