# Room with $n+k$ places and probability that $m$ places will be taken

In a room with $$n + k$$ places, $$n$$ people take random places. What is the probability that $$m \leq n$$ previously selected places will be taken?

If I understand it correct we choose $$m$$ places earlier (let's say: a chair number 1, a chair number 10, a chair number 18, a chair number 19, ... - "m" places in total) and we want to calculate the probability that all of them will be taken when "n" people take places in a room with "n+k" places.

I do not know how to start. I would be grateful for any hints.

• Hint: hypergeometric distribution. To be found is $P(X=m)$ where $X$ denotes the number of the number of previously selected places that are taken. Dec 30, 2019 at 12:50
• To clarify, can multiple people select the same "place"? When you say $m$ "previously selected places" will be taken, what do you mean here? Had the people previously been put into places and we are asking how many of those places used in the previous arrangement are used in the new arrangement? Or perhaps are you asking for the the probability that within an arrangement, exactly $m$ distinct places are used (with some of those $m$ occurring multiple times)? Dec 30, 2019 at 12:56
• I do not think that multiple people can select the same place. But to be honest I do not know anything more that I wrote above. That is how the task is formulated on the task list. Dec 30, 2019 at 13:02
• Did we choose in advance $n$ places and we want $m$ of which to be taken, or we selected $m$ and we want all to be taken?
– user706912
Dec 30, 2019 at 13:03
• If I understand it correct we somehow choose "m" places earlier (let's say: a chair number 1, a chair number 10, a chair number 18, a chair number 19, ... - "m" in total) and we want to calculate the probability that all of them will be taken when "n" people take places in a room with "n+k" places. Dec 30, 2019 at 13:12

Given a subset of the places of size $$m$$, there are $$n+k-m \choose n- m$$ ways to complete it to a subset of size $$n$$.
As each of these subsets occur with probability $${n+k\choose n}^{-1}$$, the probability of success is $$\frac{n+k-m\choose n-m}{n+k\choose k}$$
• @Jeanba $\binom{n+k}{n}=\binom{n+k}{k}$ by the same logic that $\binom{a}{b}=\binom{a}{a-b}$, to choose those which are selected is equivalent to choosing those which aren't selected. In my interpretation places could be repeated. The OP seems to disagree with that interpretation, so I retract my downvote and earlier comment. Dec 30, 2019 at 13:30
• @Uhans Given $m$ places, the numerator is the number of ways to choose $n-m$ additional places from the $n+k-m$ remaining places. Which is the number of ways to choose $n$ places that fully contain the previously selected $m$ places