= Solution
Use the positive-support <geometric distribution>, $\mathbb P(X=j)=p(1-p)^{j-1}$ for $j\geq1$, $0<p\leq1$. Differentiating the geometric series supplies
$$
\boxed{\mathbb E[X]=\frac1p,\qquad
\mathbb E[X^2]=\frac{2-p}{p^2},\qquad
\operatorname{Var}(X)=\frac{1-p}{p^2}}.
$$
For the jar, let $W_r$ be the waiting time to remove one red ball when exactly $r$ red balls remain. Every draw has success <probability> $r/n$, so $W_r$ is geometric with that parameter. The $W_r$ are independent: after each success, fresh independent draws begin a new stage, and its conditional waiting distribution depends only on the deterministic count $r$, not on earlier waiting times. Replacing a green ball leaves that stage's count unchanged.
Thus the <coupon collector problem> decomposition is $T=\sum_{r=1}^nW_r$, and
$$
\boxed{\mathbb E[T]=n\sum_{r=1}^n\frac1r},\qquad
\boxed{\operatorname{Var}(T)=n^2\sum_{r=1}^n\frac1{r^2}-n\sum_{r=1}^n\frac1r}.
$$
This is the <coupon collector waiting-time variance>. For $n=1$, the formula correctly gives zero <variance> and a deterministic one-minute completion.
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