Waiting for a Bus, arrives every 10 minutes on average What is the probability of waiting for a bus at most 30 minutes when on average a bus arrives every 10 minutes
 A: In Statistics, the number of buses arriving in a time-interval of length $t$ is usually modeled by the Poisson distribution with parameter $\lambda t$ (Assuming the buses arrive at homogeneous Poisson process, as mentioned by @Robert Israel in a comment above. This is the usual assumption that we make). $\lambda$ denotes the rate of arrival of buses in unit interval of time. Since, on average, the buses arrive every 10 minutes, $\lambda=\frac{1}{10}$.
Let $X$ denote the number of buses arriving in the time-interval, say $(0,t)$, and $T$ denote the waiting time until the first bus arrives. The crucial observation here is that $$\{X=0\} \iff \{T>t\}$$
Now the probability mass function of Poisson distribution (with parameter $\lambda t$) is given by:
$$P(X=x)=\frac{e^{-\lambda t}(\lambda t)^x}{x!}$$
where $x!=1 \times 2 \times \cdots \times x$. Since $X$ follows poisson distribution with parameter $\lambda t$, we have
$$P(T>t)=P(X=0)=e^{-\lambda t}$$
In this problem $\lambda=\frac{1}{10}$ and we are required to calculate $P(T \leq 30)$.
$$P(T \leq 30)=1-P(T>30)=1-e^{-(\frac{1}{10} \times 30)}=1-e^{-3} \approx 0.9502$$
This is the exact expression, which is close to $1$, but slightly less.
