# Probability that all people get out on the specific floor

An elevator with 5 passengers inside goes up. There are 7 floors in the building. Calculate the probability that a) each person will leave on a different floor and that b) all of them will leave on the second floor.

I know there were many questions with elevator problems, but so far I haven't found one with an answer to b) - a specific floor on which all people would leave - if I'm missing something, I'd be happy to be linked to the specific one, if not, can anyone help out? For the a) part, I think the answer would be:

$$\frac{7*6*5*4*3}{7^5}$$ Please do correct me if I'm wrong! Thank you.

For b) it is simply $$\left(\frac{1}{7}\right)^5$$

That is, each person goes off at the second floor with a probability of $\frac{1}{7}$

• Ok, thank you so much! So the number of the specific floor doesn't really matter? Because the way I see it, this is just the probability of all people getting off on the same floor, is that correct? – Jake Nov 21 '17 at 23:50
• @Jake it does matter. If it was the probability of all of them leaving on the same floor, you'd have to multiply by $7$ since there are $7$ floors where that could happen. Or: once the first person goes off on a certain floor, the probability of all the other $4$ leaving on that same floor is $(\frac{1}{7})^4$ – Bram28 Nov 21 '17 at 23:52
• So why is this not multiplied by 2, if it's the second floor? Is it because we only choose one floor and not two possible to get off on? Sorry, I really want to understand it! :) And I'll accept your answer as soon as I can. – Jake Nov 21 '17 at 23:53
• @Jake the floor's number has of course nothing to do with it: the probability of all of them leaving on the $6$-th floor is of course exactly the same as the probability of all of them leaving on the second floor. – Bram28 Nov 21 '17 at 23:55
• Thanks a lot everyone! I get it now! – Jake Nov 22 '17 at 0:05

You got (a) right.

The first person can get on any of the $7$ floors, the next person can get on any of the remaining $6$ floors, etc. giving

$$\frac{7\cdot6\cdot5\cdot4\cdot3}{7^5}\approx.1499$$

For (b)

The probability that any one person gets off on floor $2$ is $1\over{7}$ so we can just use the multiplication rule and independence to get $${1\over{7}}^5\approx5.95\cdot10^{-5}$$

If you wanted to find the probability that they all got off on the same floor (doesn't necessarily have to be floor $2$), then you would $choose$ $1$ of the $7$ floors:

$${7\choose{1}}{1\over{7}}^5 \approx4.16\cdot10^{-4}$$

Note that this value is larger since it can be any of the $7$ floors that they all get off on.