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There is a building with 3 floors, on each floor there are 4 apartments, and in each apartment lives 1 person.
GIven that 3 people meet in the entrance of the building, what is the probability they all live on different floors.

The first person has 12 apartments to choose from, the second have $\frac{8}{11}$ and the third $\frac{4}{10}$ so it is $\frac{8}{11}$*$\frac{4}{10}$=$\frac{32}{110}$

Because every floor has the same number of apartments and because the apartments are randomly chosen, is there a way to calculate the probability without using the number of apartments?

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  • $\begingroup$ The number of apartments matters. if you imagine millions of apartments per floor then you can effectively ignore the fact that you have removed one person (hence one apt) from the sample. In that case the probability that two randomly chosen people live on different floors is, approximately, $\frac{2}{3}$ and the probability that third is on yet another floor is approximately $\frac{1}{3}$. Thus the answer would be approximately $\frac{2}{9}$. $\endgroup$ – lulu Jul 7 '15 at 18:08
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If you assume you have $3$ floors and $k$ apartments on each floor, then your formula will have to involve $k$. The only thing you can do to get a formula that does not involve $k$ is to compute the limit as $k \to \infty$, which will be $(2/3)(1/3) = 2/9$.

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Well the probability for $k$ apartments in each floor is

$$\frac{k^3}{3k \choose 3} = \frac{2k^2}{(3k-1)(3k-2)}$$

which deeply depends on $k$. So you may forget about a formula not using the number of apartments...

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