# Elevator floor selection probability question

Three people get into an elevator on the main floor of a building that has 8 floors (this building has 7 floors above the ground-level). Assuming that each of the three people will get off the elevator on one of floors 2 through 8, and each does so randomly.

What is the probability that at least 2 of the three people get off at the same floor?

The only way I could think of doing it was add 1 the number of ways that exactly 2 could get off (108), so that i'm adding on the chance that all three step off at the exact same time. But that was incorrect, i'd love it if somebody could explain the logic behind what I should do.

Method 1: The complementary event is that the people exit the elevator on three different floors. Therefore, you can compute the desired probability by subtracting the probability that the people exit the elevator on three different floors from $1$.

Each person can exit the elevator on one of seven floors, so there are $7^3$ possible exit patterns. The people can exit the elevator on three different floors in $7 \cdot 6 \cdot 5$ ways. Hence, the probability that the people exit on three different floors is $$\frac{7 \cdot 6 \cdot 5}{7^3} = \frac{6 \cdot 5}{7^2} = \frac{30}{49}$$

Therefore, the probability that at least two people exit on the same floor is $$1 - \frac{30}{49} = \frac{19}{49}$$

Method 2: Your strategy of adding the probability that exactly two people get off on the same floor to the probability that all three people get off on the same floor will also work.

However, you have calculated the probability that exactly two people get off on the same floor incorrectly.

There are $\binom{3}{2}$ ways to select two people who get off the elevator on the same floor, $7$ ways to select the floor on which they exit, and $6$ ways to select the floor on which the other person exits. Hence, there are $\binom{3}{2} \cdot 7 \cdot 6 = 126$ ways for exactly two people to get off on the same floor.

Since there are $7$ ways for all three people to get off on the same floor, there are $126 + 7 = 133$ favorable cases.

Since each person could exit the elevator on one of seven floors, there are $7^3 = 343$ cases in total. Hence, the probability that at least two people exit the elevator on the same floor is

$$\frac{\binom{3}{2} \cdot 7 \cdot 6 + 7}{7^3} = \frac{133}{343} = \frac{19}{49}$$