Independent Trials: Probability of Events A_i

[IniquiTrance]

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]

 

[Martin Rattigan]

You have just proved that the events [itex]A_i[/itex] and [itex]A_j[/itex] are not independent.

If they were then the probability of no colours being chosen would be

[tex](\frac{N-1}{N})^{nN}[/itex]

but this exprssion is obviously not 0.

(Intuitively failing to draw one colour increases the chances of drawing each of the others.)

 

[IniquiTrance]

You have just proved that the events [itex]A_i[/itex] and [itex]A_j[/itex] are not independent.

If they were then the probability of no colours being chosen would be

[tex](\frac{N-1}{N})^{nN}[/itex]

but this exprssion is obviously not 0.

(Intuitively failing to draw one colour increases the chances of drawing each of the others.)

Ah ok. Thanks!