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Given:

  • There are $m$ rows of balls.
  • Each row has $k$ red balls and $n-k$ white balls, randomly placed.

With what probability do there exist $l$ totally red collumns?

visual aid

Update: The only thing I found out is the total number of possibilities:

$${n\choose k}^m$$

Carucel
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  • Remark: please show your attempts. Hint: start by calculating the total number of outcomes. – maxmilgram Jan 22 '19 at 10:45
  • @maxmilgram I only managed to calculate the total number of outcomes: ${n\choose k}^m$ – Carucel Jan 22 '19 at 11:02
  • There is no $l$ in your answer, which makes it highly suspicious. – darij grinberg Jan 22 '19 at 11:35
  • For the first row we have no restrictions. Then we have ${k\choose l}$ options to choose the red columns. For all other rows $l$ spots are already filled with red marbles leaving $k-l$ marbles to be distributed among $n-l$ places. This should give ${n\choose k}{n-l\choose k-l}^{m-1}{k\choose l}$ positive outcomes. – maxmilgram Jan 22 '19 at 12:07
  • @maxmilgram, thank you, it seems that you have calculated the outcomes for which there exists at least $l$ red collumns, right? – Carucel Jan 22 '19 at 12:22
  • You are completely right! – maxmilgram Jan 22 '19 at 12:33
  • Do you want the probability that there are at least $\ell$ red columns, or exactly $\ell$ red columns? – Mike Earnest Jan 22 '19 at 21:28

1 Answers1

1

Hint: I think the easiest way is to use the generalized principle of inclusion exclusion. Specifically, if there are several events $E_1,E_2,\dots,E_n$, and we let $X$ denote the number of these events which occur, then $$ P(X\ge \ell) = \sum_{j=\ell }^n (-1)^{j-\ell}\binom{j-1}{\ell-1}\sum_{S\subseteq [n],|S|=j} P(E_S)\tag 1 $$ where $S$ ranges over all subsets of $[n]\stackrel{\text{def}}=\{1,2,\dots,n\}$ of size $j$, and $E_S=\bigcap_{i\in S}E_i$.

Specifically, $E_i$ be the event that the $i^{th}$ column is all red. Then $P(E_S)$ is the event that all columns in $S$ are red, which is not hard to compute. Plugging that answer into $(1)$, you get a summation formula for the probability that at least $\ell$ red columns exist.

Alternatively, the probability that exactly $\ell$ columns are red is $$ P(X=\ell) = \sum_{j=\ell }^n (-1)^{j-\ell}\binom{j}{\ell}\sum_{S\subseteq [n],|S|=j} P(E_S)\tag 2 $$

See this answer for a proof of $(2)$, from which you can derive $(1)$ by summing (after applying Pascal's identity, there will be a telescopic sum).

Mike Earnest
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  • Thank you! $P(E_S)={n-|S|\choose k-|S|}^m/{n\choose k}^m$ and $|{S\subseteq[n]:|S|=j}|={n\choose j}$, which gives $P(X\geq\ell)=\sum_{j=\ell}^n(-1)^{j-\ell}{j-1\choose\ell-1}{n\choose j}{n-j\choose k-j}^m/{n\choose k}^m$. This is not simplifyable by Mathematica and not the same as the solution given by maxmilgram: ${n\choose k}{n-\ell\choose k-\ell}^{m-1}{k\choose \ell}/{n\choose k}^m$. – Carucel Jan 23 '19 at 08:37
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    @Carucel Indeed, maxmilgram's answer is incorrect. Their answer is the first summand of my answer, indicating they are not correcting for overcounting; their answer overcounts cases where there are more than $\ell$ red columns, as they are not uniquely represented by choosing $\ell$ red columns, then choosing the rest of the entries arbitrarily. – Mike Earnest Jan 23 '19 at 18:06