Say I have an urn containing a set of balls numbered(adsbygoogle = window.adsbygoogle || []).push({}); 1throughN. I then makenselections from the urn with replacement. Thus each selection is independent.

Let the event: [tex]A_{i}[/tex]

[tex]i=1,2,3....,N[/tex]

be that the ball numberedwas not chosen in theinselections.

Then:

[tex]P(\mathbf{A_{i}})=\left(\frac{N-1}{N}\right)^{n}[/tex]

and [tex]P(\mathbf{A_{i}A_{j}})=\left(\frac{N-2}{N}\right)^n}[/tex] and so on....

But aren't the events [tex]A_{i}[/tex] [tex]i=1,2,3....,N[/tex] independent as well?

So shouldn't the probability of

[tex]P(\mathbf{A_{i}A_{j}})=P(\mathbf{A_{i}})P(\mathbf{A_{j}})=\left(\frac{N-1}{N}\right)^{2n}=\left(\frac{2(N-1)}{N}\right)^{n}[/tex]

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# Independent trials

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