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In hypergeometric distribution the following info is given to us. There are $N$ objects out of which $r$ objects are desirable and $N-r$ undesirable. To select $x$ objects from $r$ and $n-x$ objects from $N-r$ we apply hypergeometric distribution function.

My question is: do we pick those $x$ objects out together or do we pick the objects one by one? The formula seems to indicate that $x$ objects are picked together and not one by one. But that doesn't go well with the fact that the sample space changes with each selection because of no replacement of objects once selected...

user90596
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    It is not really necessary to work with a changing sample space. What we do instead is making use of conditional probabilities. In the example I gave in my answer we state that e.g $P(G_1\cap G_2\cap B_3)=P(G_1)P(G_2\mid G_1)P(B_3\mid G_1\cap G_2)=\frac68\frac57\frac26$ is the probability on "green, green, blue" in this order. – drhab Sep 25 '18 at 07:14

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Actually there is no essential difference between choosing $x$ objects one by one without replacement and choosing them at the same time. Sometimes it can be useful to practicize some order of selection.

If e.g. there are $8$ balls of which $6$ are green and $2$ are blue and $3$ are selected then what is the probability that exactly $2$ are green?

This can be answered by: $$\frac{\binom62\binom21}{\binom83}$$ in the scenario of taking $3$ balls together and applying the formula you have in mind.

But also by: $$3\times\frac68\frac57\frac26$$ in the scenario of taking $3$ balls one by one. Here factor $3$ arises because there are $3$ equiprobable possibilities: $GGB$,$GBG$ and $BGG$.

drhab
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