Let’s once again test a series of critical system components until you find two that fail. Again, suppose a particular component has a p=0.01 probability of breaking on any given trial. Since you will stop when you encounter the third failure, you can model this situation with a negative binomial distribution and the probability that the 2nd component fails on the x-th trial is given by
\begin{equation*}
f(x) = \binom{x-1}{2-1} 0.99^{x-2} \cdot 0.01^2 = (x-1) \cdot 0.99^{x-2} \cdot 0.01^2.
\end{equation*}
Once again, let’s compute the likelihood that you get the second failure on one of the first five trials. Then,
\begin{equation*}
F(5) = f(2)+f(3)+f(4)+f(5) \\ = 0.01^2 + 2 \cdot 0.99 \cdot 0.01^2 + 3 \cdot 0.99^2 \cdot 0.01^2 + 4 \cdot 0.99^3 \cdot 0.01^2 \approx some nice number
\end{equation*}
which is still relatively small.