Probability of intersection from multiple sampling of the same population Here is an example case:


*

*I have a population of 10,000 items. Each item has an unique id.  

*I randomly pick 100 items and record down the ids  

*I put the 100 items back into the population  

*I randomly pick 100 items again, record down the ids and replace.  

*In total, I repeat this random sampling 5 times


What is the probability that $X$ number of items appear in all 5 random samplings?
I am not very well versed in statistics. Would this be correct for $X = 10$?


*

*For each sampling, the number of possible combinations of 100 items from 10,000 is ${\rm binom}(10000, 100)$

*Out of all possible combinations of 100 items, ${\rm binom}(9990, 90) * {\rm binom}(100, 10)$ combinations contain 10 specific items

*The probability of having 10 specific items is $({\rm binom}(9990, 90) * {\rm binom}(100, 10)) / {\rm binom}(10000, 100)$

*The calculated probability to the power of 5 would represent 5 indepenent samplings. 


So essentially we are just calculating 5 independent hypergeometric probabilities and then multiplying them together? I feel like I am missing a step somewhere.
 A: I just ran into a similar problem and, even though I also don't know if this is the correct solution, approached it like this:
You are interested in the occurrence of $X$ items in 5 samples á $100$ items of $10,000$ items total. You could think of an urn with $X$ white balls and $10,000-X$ black balls. $100$ balls are taken out and $p_h$ is the probability that you have all $X$ white balls in your set. If you do this $5$ times (independently), I would multiply it: $p = {p_h}^5$.
I could even think of one step further and wrap it around the binomial distribution: If you have a coin which comes up head with probability $p_h$ (the probability that you have all items in your set) and you toss it $5$ times, what is the probability of getting $5$ heads? $p = {5\choose 5}{p_h}^5 (1-{p_h})^{5-5} = {p_h}^5$.
A: 
What is the probability that $X$ number of items appear in all 5 random samplings?

Building on what Hans said, you want to always get the same $X$ ids in each sample of 100 and 100-$X$ ids from among the remaining 10000-$X$. The probability of doing so for a given sample is given by the hypergeometric function for $X$ successes in a draw of 100 from a population of 10000 with $X$ possible success states: $P = \frac{{X \choose X}{10000-X \choose 100-X}}{10000 \choose 100}$. For 5 samples, you would take $P^5$.
However, that we presuppose knowing the $X$ ids that are shared, and there are $10000 \choose X$ ways to select those $X$ ids. So your final answer would be ${10000 \choose X} P^5$.
