# Independent trials

1. Apr 15, 2010

### 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
$$P(\mathbf{A_{i}A_{j}})=P(\mathbf{A_{i}})P(\mathbf{A_{j}})=\left(\frac{N-1}{N}\right)^{2n}=\left(\frac{2(N-1)}{N}\right)^{n}$$

Last edited: Apr 15, 2010
2. Apr 16, 2010

### 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

[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.)

3. Apr 16, 2010

### IniquiTrance

Ah ok. Thanks!