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Quite new in stats... definitely not my strong area. I came across this probability question, and I am not sure how to do this!

The question goes:

pretend that there's this meeting scheduling engine used by this company and is not synced in real time, so when people schedule their meetings online to book a room, there may be overlaps. let's say there are $N$ rooms, and $M$ meetings, where $M \leq N$, what is the probability that all meetings scheduled to a different room?

My thought was that, the first meeting doesnt matter, can be in any room; then the 2nd meeting has $\frac{1}{N-1}$ chance of being in a room. so for two rooms not colliding, the chance of them being in separate rooms is $\frac{1}{N-1}$. Right? I am not confident about this one neither...

Any hint/advice/guidance helps!

update to clarify:

1) each room can only host up to one meeting
2) one meeting can only happen in one room

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    It depends on how the meetings are scheduled. Can two meetings be scheduled for the same room? – 2017-01-06
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    Hi @copper.hat, no, as stated in the title, all the meetings are scheduled to a different room. so no 2 meetings in a same room. hope that clarifies. – 2017-01-06

3 Answers 3

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Same solution as the others with a slightly different perspective (somewhat in line with the reasoning you tried to provide).

Let $E_i$ be the event that room $i$ does not collide with rooms $1,2,3,...,i-1$.

As you noted the first room does not matter.

$$P(E_1)=1$$

For the second room, there are $N-1$ rooms we can pick out of the $N$ total taking into account the first room was chosen:

$$P(E_2|E_1)=\dfrac{N-1}{N}$$

Similarly,

$$P(E_3|E_2,E_1)=\dfrac{N-2}{N}$$

And the pattern continues in this fashion.

$$P(E_i|E_{i-1},E_{i-2},\cdots ,E_1) = \dfrac{N-i+1}{N}$$

You seek

$$P(E_1\cap E_2\cap \cdots\cap E_m) = P(E_1)P(E_2|E_1)P(E_3|E_2,E_1)\cdots P(E_m|E_{m-1},E_{m-2},\cdots,E_1)$$

$$=1\cdot \dfrac{N-1}{N}\cdot \dfrac{N-2}{N}\cdots \dfrac{N-m+1}{N}$$

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    ah, your answer totally make sense to me! thanks for going off my thought process, its easier for me to follow. can i ask why you use (N-M+1) as the last element for the numerator? why +1? – 2017-01-06
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    Notice for $E_2$ the numerator was not $N-2$ but $N-2+1$ or $N-1$. And the numerator for $E_3$ was $N-3+1$ or $N-2$. When we are choosing the $m^{th}$ room, there are only $m-1$ taken up so there are $N-(m-1)$ rooms left to choose from. $N-(m-1)=N-m+1$. – 2017-01-06
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    omg thank you so much! okay that totally make sense now. thank you very very much! i missed the pattern – 2017-01-06
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The total number of ways to distribute meetings to rooms is $N^M$, because for example the first meeting can take place in $N$ rooms, the second also in $N$ rooms, and so on.

Let's calculate the number of favorable situations. The first meeting has $N$ choices. The second has $N-1$ choices and so on. So the probability of all meetings are in different rooms is $$\frac{N(N-1)\cdots(N-M+1)}{N^M}$$

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    Hi msm, thanks so much for trying to help! i misinterpreted your answer. my apology. i do have a question regarding the N^M, how did you come up with that? – 2017-01-06
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    You want to assign rooms to meetings. For the first meeting there are $N$ options. For the second meeting there are also $N$ options. So for each of the $M$ meetings there are $N$ options, which means $N^M$ choices. – 2017-01-06
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    ah okay. that make sense! but why did you use (N-M+1) as the last element for the numerator? why +1? – 2017-01-06
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    The first term ($N$) is for the first meeting. $N-1$ for the second meeting, $N-2$ for the third meeting. So for the $M$'th meeting the corresponding term is $N-(M-1)=N-M+1$ – 2017-01-06
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    so your answer is basically N!/(N^M)? – 2017-01-06
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    No. if you want to write it using factorial, it is $$\frac{N!/(N-M)!}{N^M}$$ – 2017-01-06
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51267/discussion-between-alwaysaskingquestions-and-msm). – 2017-01-06
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If you are not talking about scheduling (for which time overlap would an issue), but instead talking about just where the meeting takes place, then there are $N^M$ possible assignments of meeting rooms. There are $\binom{N}{M}$ ways to choose $M$ different rooms for the meetings and $M!$ ways to assign the $M$ meetings to those rooms, so the probability is ${\binom{N}{M}M! \over N^M}$.

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    You need to multiply that by $M!$ to account for the ways the meetings can be permuted amongst the $M$ rooms chosen. – 2017-01-06
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    Thanks! I realized that but my train arrived before I had a chance to edit, stuck on phone for now. – 2017-01-06
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    If you like, I can make a quick edit; just let me know. – 2017-01-06
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    That would be much appreciated, otherwise I will get to in an hour and a bit. Thanks! – 2017-01-06
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    Done! (And now +1.) – 2017-01-06
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    @BrianM.Scott: Much appreciated! – 2017-01-06
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    can i please ask why multiply by M! ? i thought the (N M) will take care of assigning M meeting rooms to N total rooms already? – 2017-01-06
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    @alwaysaskingquestions: That was my original omission, which Brian fixed. There are $\binom{N}{M}$ ways of choosing $M$ rooms, but this does not take into account the $M!$ ways of assigning these rooms. – 2017-01-06
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    @copper.hat but (N M) already take into account when M rooms are ordered differently right? it calculates all the possible ways of choosing M rooms from N rooms total. i am not seeing how its different from M! ways of choosing the rooms? (thanks for being patient with me!) – 2017-01-06
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    @alwaysaskingquestions: No. At an extreme, there is exactly one way of choosing $N$ out of $N$ rooms, but there are $N!$ ways of assigning $N$ meetings to $N$ rooms. – 2017-01-06
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    @copper.hat ohhh i see. b/c (N M) does not count the orders of assigning the rooms. am i right? okay that make sense now! – 2017-01-06