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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.

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  • $\begingroup$ 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. $\endgroup$
    – drhab
    Dec 30, 2019 at 12:50
  • $\begingroup$ 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)? $\endgroup$
    – JMoravitz
    Dec 30, 2019 at 12:56
  • $\begingroup$ 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. $\endgroup$
    – Uhans
    Dec 30, 2019 at 13:02
  • $\begingroup$ 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? $\endgroup$
    – user706912
    Dec 30, 2019 at 13:03
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    $\begingroup$ 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. $\endgroup$
    – Uhans
    Dec 30, 2019 at 13:12

1 Answer 1

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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 $$\frac{n+k-m\choose n-m}{n+k\choose k}$$

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    $\begingroup$ I do not understand how you interpret the problem... I assumed that the first person chooses an empty place then the second person etc... $\endgroup$
    – user706912
    Dec 30, 2019 at 13:00
  • $\begingroup$ @AsafRosemarin I interpreted the problem exactly as you did. EDIT: I think there is a typo though, it should be n instead of k in the denominator. $\endgroup$
    – Jeanba
    Dec 30, 2019 at 13:22
  • $\begingroup$ Each one chooses an empty place. I do not understand the formula in the numerator. Could you explain a bit more? I edited the question a little - I understand it so that "m" places are choosen eariler and we check in how many combinations all of them are taken. $\endgroup$
    – Uhans
    Dec 30, 2019 at 13:27
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    $\begingroup$ @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. $\endgroup$
    – JMoravitz
    Dec 30, 2019 at 13:30
  • $\begingroup$ @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 $\endgroup$
    – user706912
    Dec 30, 2019 at 13:40

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