= Solution
For the <geometric distribution> supported on positive integers, set $q=1-p$. The geometric series and its derivative give
$$
E(T)=\sum_{t=1}^{\infty}tpq^{t-1}
=p\frac{d}{dq}\left(\frac1{1-q}\right)
=\frac p{(1-q)^2}
=\boxed{\frac1p}.
$$
Equivalently, the sum of the tail probabilities is $\sum_{k\geq0}P(T>k)=\sum_{k\geq0}q^k=1/p$. For $p=0.02$, the expected wait is \b[50 years], so this is the \b[50-year <return level>]. The exceedance probability determines the return period, but does not by itself determine a numerical flow rate.
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