What is the probability of waiting for a bus at most 30 minutes when on average a bus arrives every 10 minutes
[Math] Waiting for a Bus, arrives every 10 minutes on average
probability
Related Solutions
As mentioned in the comments, the answer depends very much on the model used to describe the passage times of the buses. The deterministic situation where the passage times of buses of type $k$ are $s_k+m_k\mathbb N$ for some initial passage time $s_k$ in $(0,m_k)$ is too unwieldy to be dealt with in full generality hence we now study two types of assumptions.
(1) Fully random passage times
Here the passage times of buses of type $k$ are a Poisson process of intensity $1/m_k$ and the passage times of buses of different types are independent. Then, starting at time $t_0$, the next bus of type $k$ arrives after a random time exponential with mean $m_k$ hence the waiting time $T$ is such that $$ \mathbb P(T\gt t)=\prod_k\mathbb P(\text{no bus of type}\ k\ \text{in}\ (t_0,t_0+t))=\prod_k\mathrm e^{-t/m_k}=\mathrm e^{-t/m}, $$ where $$ \frac1m=\sum_k\frac1{m_k}. $$ In particular, $T$ is exponentially distributed with parameter $1/m$, hence $$ \mathbb E(T)=m. $$ The case $m_1=m_2=\cdots=m_n$ yields $$ \mathbb E(T)=\frac{m_1}{n}. $$ (2) Fully periodic passage times with random uniform initializations
Here, buses of type $k$ pass at times in $S_k+m_k\mathbb N$ where $S_k$ is uniform on $(0,m_k)$ and the random variables $(S_k)$ are independent. Now, starting at time $t_0$, the next bus of type $k$ arrives after time $t_0+t$ if $t\leqslant m_k$ and if $S_k$ is not in a subinterval of $(0,m_k)$ of lenth $t/m_k$. Thus, $$ \mathbb P(T\gt t)=\prod_k\left(1-\frac{t}{m_k}\right),\qquad t\leqslant \bar m=\min\limits_km_k. $$ A consequence is that $$ \mathbb E(T)=\int_0^{+\infty}\mathbb P(T\gt t)\,\mathrm dt=\int_0^{\bar m}\prod_k\left(1-\frac{t}{m_k}\right)\,\mathrm dt. $$ Expanding the product yields $$ \mathbb E(T)=\sum_{i\geqslant0}(-1)^i\bar m^{i+1}\frac1{i+1}\sum_{|K|=i}\frac1{m_K}, $$ where, for every subset $K$, $$ m_K=\prod_{k\in K}m_k. $$ For example, time intervals $m_1$, $m_2$, $m_3$ with minimum $m_1$ yield $$ \mathbb E(T)=m_1-\frac{m_1^2}2\left(\frac1{m_1}+\frac1{m_2}+\frac1{m_3}\right)+\frac{m_1^3}{3}\left(\frac1{m_1m_2}+\frac1{m_2m_3}+\frac1{m_3m_1}\right)-\frac{m_1^4}{4m_1m_2m_3}, $$ which can be simplified a little bit (but not much) into $$ \mathbb E(T)=\frac{m_1}2-\frac{m_1^2}{6m_2}-\frac{m_1^2}{6m_3}+\frac{m_1^3}{12m_2m_3}. $$ The case $m_1=m_2=\cdots=m_n$ yields $$ \mathbb E(T)=\frac{m_1}{n+1}. $$
The expected waiting time = 11.25 minutes, in my opinion.
Expected time if the inter-arrival time is 15 minutes with probability 1 = 7.5 mins
Expected time if the inter-arrival time is 30 minutes with probability 1 = 15 mins
Now, expected waiting time = $0.5 * (7.5 + 15)=11.25$
Best Answer
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.