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I am currently working on the following question: Suppose you have a finite population $P$ of size $N$. We select a sample $S_1$ using a random sampling without replacement of size $n_1$. Then we select a sample $S_2$ from $P-S_1$ using random sampling without replacement of size $n_2$.

Then we define $S= S_1 \cup S_2$. What is the sampling distribution of $S$?

$My \ answer$: There are $\binom{N}{n_1}$ possible ways to have a random sample $S_1$. Then there are $\binom{N-n_1}{n_2}$ possible ways to have a random sample $S_2$. Hence the number of possibilities for a random sample $S$ equals $\binom{N}{n_1}\binom{N-n_1}{n_2}$. Now we have counted possible outcomes multiple times. There are $\binom{n_1+n_2}{n_1}$ possible ways to have a specific selection. Hence the total numer of outcomes equals $\binom{N}{n_1}\binom{N-n_1}{n_2}/\binom{n_1+n_2}{n_1} = \binom{N}{n_1+n_2}$.

Therefore the probability of a specific sample equals $[\binom{N}{n_1+n_2}]^{-1}$

Is my reasoning correct? Thanks in advance!

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  • $\begingroup$ What do y0u mean by sampling distribution ? $\endgroup$ – Satish Ramanathan Nov 18 '17 at 11:37
  • $\begingroup$ If you take $N=3, n_1=n_2=1$ then $\binom{N}{n_1+n_2} = \binom{3}{2} = 3$. However $\binom{N}{n_1}\binom{N-n_1}{n_2} = \binom{3}{1}\binom{2}{1} =6$. $\endgroup$ – user421927 Nov 18 '17 at 12:15
  • $\begingroup$ @kees1 You are correct. I will delete my former comment. $\endgroup$ – drhab Nov 18 '17 at 13:32
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Yes, but more directly you could say that this probability equals:$$\binom{N}{n_1+n_2}^{-1}$$

There is no essential difference between taking two consecutive samples of sizes $n_1$, $n_2$ respectively that are afterwards joined and one sample of size $n_1+n_2$.

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  • $\begingroup$ Thanks for your response. I noticed that I did not take into account that we can obtain the same random sample. In my previous answer I counted several samples more then once. I have changed my reasoning. $\endgroup$ – user421927 Nov 18 '17 at 12:00

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