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: A_{i}
i=1,2,3...,N

be that the ball numbered i was not chosen in the n selections.

Then:

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

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

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

So shouldn't the probability of
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}

 

[Martin Rattigan]

You have just proved that the events A_i and A_j are not independent.

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

(\frac{N-1}{N})^{nN}[/itex]<br /> <br /> but this exprssion is obviously not 0.<br /> <br /> (Intuitively failing to draw one colour increases the chances of drawing each of the others.)

 

[IniquiTrance]

You have just proved that the events A_i and A_j are not independent.

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

(\frac{N-1}{N})^{nN}[/itex]<br /> <br /> but this exprssion is obviously not 0.<br /> <br /> (Intuitively failing to draw one colour increases the chances of drawing each of the others.)

<br /> <br /> Ah ok. Thanks!