- #1
IniquiTrance
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Say I have an urn containing a set of balls numbered 1 through N. I then make n selections 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 numbered i was not chosen in the n selections.
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]
Let the event: [tex]A_{i}[/tex]
[tex]i=1,2,3...,N[/tex]
be that the ball numbered i was not chosen in the n selections.
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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