presented with n identical balls, one with a prize in it. Picks each ball out idependently one at a time till gets prize. I need to find the mean and variance of the number of balls needed to pick out the prize, if the balls are not replaced.
I can work it out with replacement, and i have been told to not work it out with hypergeometric series.
I know that there are going to be $\frac{n!}{2(n-2)!}$ combinations due to the formula for unordered non replacment.
I'm thinking so far it has to also do with the series from $k= 1 \ldots n$ of $\frac{1}{k} + \frac{1}{k}\frac{1}{(k-1)} + \frac{1}{k}\frac{1}{(k-1)}\frac{1}{(k-2)}$ etc but I can't think of a way to neaten that series either.
Please help!
Let $X$ be the random variable equal to the number of balls that must be picked to win the prize.
As long as the winning ball has not been picked, the probability to get the winning ball at the $k$-th pick is the product of the probabilities not to win at the first pick, at the second pick, &c., at the $k-1$-th pick, and finally to win at the $k$-th pick. As there is one less ball after each pick, using the formula of composed probabilities, this gives: $$P(X=k)=\frac{n-1}{n}\cdot\frac{n-2}{n-1}\dotsm\frac{n-k+1}{n-k+2}\cdot\frac1{n-k+1}=\frac1n$$ Thus the law of $X$ is the uniform law, and it is known that: $$E(X)=\frac{n+1}2,\quad V(X)=\frac{n^2-1}{12}.$$