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May 10, 2020 at 15:53 history edited Max Alekseyev CC BY-SA 4.0
typo corrected
May 10, 2020 at 15:22 comment added Carlos A. Astudillo Trujillo I tested your solution of selection with replacement and it is more accurate than my previous suggestion of considering $m^{'}$ and selection without replacement. Thank you again.
May 10, 2020 at 14:30 comment added Carlos A. Astudillo Trujillo Another option to answer my initial question is just to consider $m$ as $m^{′}$ in your formula without replacement and calculate $m{′}=N{\cdot}k\left(1−B_{0}(n,p)\right)$, where $B_{r}(n,p)$ is the binomial distribution with parameters $n=m$ (as defined in the initial question), $p=\frac{1}{N{\cdot}k}$ and $r$ is the number of times an element in $S_k$ is selected. Thus, $1−B_{0}(n,p)$ is the probability that an element of $S_{k}$ is selected at least once in $n$ trials.
May 10, 2020 at 14:21 vote accept Carlos A. Astudillo Trujillo
May 10, 2020 at 13:43 comment added Max Alekseyev @CarlosA.AstudilloTrujillo: Yes. Selection with replacement is similar - added it now
May 10, 2020 at 13:42 history edited Max Alekseyev CC BY-SA 4.0
added 150 characters in body
May 10, 2020 at 13:27 comment added Carlos A. Astudillo Trujillo Thank you @Max for your answer. It seems that in your solution $m$ is the number of unique selected elements, is it right?
May 10, 2020 at 5:15 history edited Max Alekseyev CC BY-SA 4.0
simplified
May 10, 2020 at 5:08 history undeleted Max Alekseyev
May 10, 2020 at 5:02 history edited Max Alekseyev CC BY-SA 4.0
corrected
May 10, 2020 at 4:40 history deleted Max Alekseyev via Vote
May 10, 2020 at 4:37 history edited Max Alekseyev CC BY-SA 4.0
added 9 characters in body
May 10, 2020 at 4:25 history answered Max Alekseyev CC BY-SA 4.0