A jar contains $n$ balls. Initially, all of the balls are red. Every minute, a ball is drawn at random from the jar, and then replaced with a green ball. Let $T$ be the number of minutes until the jar contains only green balls. Show that the expected value of $T$ is $n\sum_{i=1}^n1/i$. What is the variance of $T$?
It is presented in the context of a geometric distribution, however I am not sure I know how to use it properly. The way I started was:
$ P(T=n) = \frac{n-1}{n}\cdot\frac{n-2}{n}\cdot\dots\cdot\frac{1}{n} = \frac{n!}{n^n}$
$P(T=n+1|G=j) = \frac{n-1}{n}\cdot\dots\cdot\frac{j}{n}\cdot\frac{n-j}{n}\dots\cdot\frac{1}{n} = \frac{n!}{n^n}\cdot\frac{j}{n}$, where $G=j$ means we picked a green when there were $j$ green inside.
Hence,
$P(T=n+1) = \sum_{j=1}^{n}P(T=n+1|G=j)\cdot P(G=j) = \frac{n!}{n^n}\cdot\frac{n(n+1)}{2}$.
However, for $T\geq n+2$, I am not sure I can use the same logic. Hence I am a bit stuck. Any solution, the shorter and more interesting the better, would be appreciated.
Yes, a good rule of thumb is 'don't fight it'...i.e. don't fight the way it is being presented using geometric distributions.
The key leap is to let $T_i$ denote the number of minutes between picking the $(i-1)^{th}$ red ball and the $i^{th}$ red ball. At any minute after the $(i-1)^{th}$ red ball has been picked, there is a probability $(n-i+1)/n$ of picking a red ball the next (i.e. the $i^{th}$) red ball. So we have a geometric distribution because $$ \mathbb{P}\bigl( \{T_i = k\bigr\}) = \Bigl(1 - \frac{n-i+1}{n}\Bigr)^{k-1} \Bigl( \frac{n-i+1}{n}\Bigr) $$ It is clear then that $$ \mathbb{E}( T_i ) = \frac{n}{n-i+1}. $$ Now use the fact that $T = T_1 + \dots + T_n$.