To be complete I'll just post everything about the problem.

Q: Each brand of candy bar has one coupon in it. There are n different coupons in total; getting at least one coupon of each type entitles you to a prize. Each candy bar you eat can have any one of the coupons in it, with all being equally likely. Let X be the (random) number of candy bars you eat before you have all coupons. What are the mean and variance of X?

A. So the expected value of X is a sum of geometric discrete random variables.

$\displaystyle X_1 = geom(1)$

so you have one coupon on the first trial.

let $\displaystyle X_2$ denote the number of bars needed to get the 2nd coupon

$\displaystyle E[X_2] = geom(\frac{n-1}{n}) = \frac{n}{n-1}$

$\displaystyle E[X_n] = geom(\frac{1}{n}) = n $

$\displaystyle E[X] = E[\sum_{i=1}^{n} X_i] = \sum_{i=1}^{n} E[X_i] $

The part Im having trouble on is the variance. From my definition:

$\displaystyle Var(X) = E[(X-E[X])^2] $

so does that mean...

$\displaystyle Var(X) = E[(\sum_{i=1}{n} X_i - \sum_{i=1}{n}E[X_i])^2] $

Any help is appreciated