Often one will test a critical system component until it fails to see how long the component works. Suppose you have a particular component that on any given trial has a p=0.01 probability of breaking. You also might find it reasonable to presume that succesive trials are independent which could be the case if the component shows no wear from trial to trial. So, you can model this situation with a geometric distribution and the probability that the component fails on the x-th trial is given by
\begin{equation*}
f(x) = 0.99^{x-1} \cdot 0.01.
\end{equation*}
For some reason, you might be interested in whether the component fails on the 5th trial. The probability of this outcome is given by
\begin{equation*}
f(5) = 0.99^4 \cdot 0.01 \approx 0.0096
\end{equation*}
so it is unlikely that the component will fail on exactly the 5th trial. However, what about failing on one of the first five trials? Then, you would need
\begin{equation*}
F(5) = f(1)+f(2)+f(3)+f(4)+f(5) \\ = 0.01 + 0.99 \cdot 0.01 + 0.99^2 \cdot 0.01 + 0.99^3 \cdot 0.01 + 0.99^4 \cdot 0.01 \approx 0.049
\end{equation*}
which is still relatively small. Indeed, with such a small probability of failure, you might expect the component to last for some time. Indeed, we will uncover a formula below for the number of trials, on average, you might expect before failure.