Firstly, I understand that this question may be simply carried using the Poisson distribution. However, I attempted this question using the Exponential distribution, but am not getting the correct answer.
Firstly, I let $N$ represent the number of buses that arrive in the span of a $\textbf{minute}$. So,
$$
N \sim \text{Poisson}(\mu = \frac{9}{60} = 0.15)
$$
Then, let $T$ represent the time (in minutes) between each bus. So,
$$
T \sim \text{Exponential}(\mu = \frac{1}{\lambda} = \frac{1}{0.15} = \frac{20}{3})
$$
As the Exponential distribution is "memory-less", the probability of a bus arriving after 20 minutes given that one has just passed, is just the probability of a bus arriving after 20 minutes.
So, the probability of needing to wait more than 20 minutes given that one has just passed can be calculated as
$$
\begin{align}
P(T > 20) &= 1 – P(T \leq 20) \\
&= 1 – F(20) \text{ where $F$ is the CDF of $T$} \\
&= 1 – (1 – e^{-\lambda t}) \tag{ CDF formula for exponential dist} \\
&= e^{-\frac{20}{3} \cdot 20} \\
&= e^{-\frac{400}{3}}
\end{align}
$$
However, the answer should simply be $e^{-3}$.
Let $X$ be the number of buses that arrive in 20 minutes. Then,
$$
X \sim \text{ Poisson}(\frac{1}{3} \cdot 9 = 3)
$$
The probability of needing to wait for more than 20 minutes is same as the probability if no buses arrived within 20 minutes, so we can solve the question as
$$
\begin{align}
P(X = 0) &= \frac{\lambda^{x}e^{-\lambda}}{x!} \\
&= \frac{3^{0}e^{-3}}{0!} \\
&= e^{-3}
\end{align}
$$
If anyone would know where I went wrong in my working in using the exponential distribution, I would be extremely grateful.
Best Answer
The probability of no bus arriving in time $t$ is $e^{-\lambda t}$ whether you use an exponential distribution or a Poisson distribution, if the rate of arrival is $\lambda$.