Let’s consider a simple example for rolling a 24-sided die until you get a multiple of 9...that is, either a 9 or an 18. Successive rolls of a die would appear to be independent events and the probability of getting a 9 or 18 on any given roll is \(p = \frac{1}{12}\text{.}\) What is then the likelihood that it takes more than three rolls in order for you to get your first success?
This is easily modeled by a geometric distribution and you are looking for
\begin{equation*}
P(X > 3) = 1 - P(X \le 3) = 1 - F(3) = 1 - f(1) - f(2) - f(3) \\
1 - \frac{1}{12} - \frac{11}{12} \cdot \frac{1}{12} - (\frac{11}{12})^2 \frac{1}{12} \approx 0.77025.
\end{equation*}