Originally Posted by

**meymathis** My apologies for some of the confusion (I'm new at the forums and didn't see your "thanks"). I thought people were arguing for the binomial distribution. Looking more carefully (which I should have done before) at awkward's example, the main thrust of my argument still holds, I think. Probably the most succinct way of saying it is this:

$\displaystyle Pr(X_i=0) = (1-1/n)^k \text{ for } i = 1,2, \dots ,n$

doesn't work because the number of people on the bus is not constantly $\displaystyle k$.

$\displaystyle \mathbf{P}(X_2=0|X_1=1)\neq$$\displaystyle \mathbf{P}(X_2=0|X_1=0)$ because there are no longer *k* people on the bus, then

$\displaystyle Pr(X_2=0|X_1=1) = (1-1/n)^{k-l}$

where $\displaystyle l$ are the number of people that got off at the first stop.

Again, sorry for not being clearer before.