# Elevator Probability Question

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

a) all exit at different floors

b) all exit at the same floor

c) two get off at one floor and two get off at another

For a) I found $4!$ ways for the passengers to get off at different floors, so $$\frac{4!}{4^4} \text{would be the probability} = \frac{3}{32}$$

For b) there are only four ways for them to all exit on the same floor, so $$\frac{4}{256} = \frac{1}{64}$$

For c) am I allowed to group the $4$ people so that I am solving for $2$ people technically? For two people there would be $12$ possibilities, and there are three ways to group the $4$ individuals, so $$\frac{12 \cdot 3}{256} = \frac{9}{64}$$

I'm not sure if I'm doing these right, can you please check? Thank you.

• For the third, I would probably argue thus: the two floors can be chosen in $\binom{4}{2}$ ways. For each way, the $2$ people who get off at the lower of these floors can be chosen in $\binom{4}{2}$ ways, for a total of $\binom{4}{2}\binom{4}{2}$ (same result as yours). – André Nicolas May 8 '14 at 16:51

First we select two of the four floors, which can obviously be done in $\dbinom{4}{2} = 6$ ways. Now we select two people from the four people to leave on the first floor, which can be done in $\dbinom{4}{2} = 6$ ways. We see that there are a total of $6 \times 6 = 36$ successful outcomes.
There are a total of $4^{4} = 256$ ways. Our probability is thus $\frac{36}{256} = \boxed{\frac{9}{64}}.$
Your calculation for $c)$ is correct, though, as the comments and answers show, you seem not to have expressed it in a readily comprehensible manner.
I would express what I understand your argument to be as: There are $3$ ways to split the $4$ people into $2$ groups. Then we can treat those two groups like $2$ people and choose the floors they get off at, $4$ options for the first group and $3$ for the second, or $12$ overall, for a total of $3\cdot12=36$ options out of $4^4=256$.
a and b are correct. For c, I would say there are ${4 \choose 2}$ ways to choose the first pair, $4$ ways to choose the floor they get off on, and $3$ ways to choose the floor the other pair gets off on, but we need to divide by $2$ because we can swap the pairs. That gives $\frac {6\cdot 4 \cdot 3}{2 \cdot 4^4}=\frac 9{64}$ This is the same as your answer, but I don't understand the logic that gives $12$ possibilities for a pair. We don't care about the order of selection in a pair.