Here is a problem I am currently thinking about:
An airport shuttle makes tours to transport passengers from a train station to two possible terminals, $A$ and $B$. After carrying all of its passengers to their destination terminals, the shuttle returns to the train station to pick up the next passengers. The tour of the shuttle takes:
$-$12 minutes if the shuttle is carrying passengers headed to both terminals.
$-$10 minutes if the shuttle is carrying passengers headed to terminal $B$ but none to terminal $A$.
$-$8 minutes if the shuttle is not carrying any passengers headed to terminal $B$ (including if there are no passengers at all).
New passengers headed to terminal $A$ keep arriving at the train station with independent waiting times which follow $\text{Exp}(1/2)$ distribution, and new passengers headed to terminal $B$ arrive with independent waiting times which follow $\text{Exp}(1/6)$ distribution, independently also of each other. Always returning from its previous tour, the shuttle picks up all new passengers that have arrived meanwhile, and starts a new tour to carry them to their destination terminals.
(a) Construct a discrete time three state Markov chain describing the tours of the shuttle, such that the states indicate whether the tour has passengers headed to both terminals, has passengers headed to terminal $B$ but not to terminal $A$, or only has passengers headed to terminal $A$ (including the case of no passengers at all). Give the transition probability matrix of the Markov chain.
(b) In the long run, what is the average duration of the tours of the shuttle?
We're given that passengers arrive with independent waiting times, but what does this mean exactly? Do they mean to say that passengers arrive with independent interarrival times? Also, in part (b) we're asked to evaluate the average duration of the tours in the shuttle. This is slightly confusing to me because the duration of the shuttle is fixed and not random. Any insight would be greatly appreciated.
Here is my attempted solution. I am assuming the question means to say passengers arrive with independent interarrival times. Let state $I$ denote the event that shuttle contains passengers travelling to both terminals, state $II$ the event the the shuttle contains passengers travelling to terminal $B$ but not $A$, and state $III$ the event that the shuttle doesn't contain any passengers from terminal $B$. Let's first assume that the shuttle is in state $I$, meaning it's on a 12 minute trip containing passengers headed to both terminals. Take $Y_A,Y_B$ as the random variables which count the number of passengers who arrive within this 12 minute period travelling to terminal $A$ and terminal $B$, respectively. Notice how $Y_A\sim \text{Poisson}\bigg(12\cdot \frac{1}{2}\bigg)$ whereas $Y_B\sim \text{Poisson}\bigg(12\cdot \frac{1}{6}\bigg)$. If $X_i\in\{I,II,III\}$ denotes the state of the shuttle on the $i^{\text{th}}$ trip, then $$P(X_{i+1}=I|X_i=I)=P(Y_A>0,Y_B>0)=(1-e^{-6})(1-e^{-2})$$ $$P(X_{i+1}=II|X_i=I)=P(Y_A=0,Y_B>0)=e^{-6}(1-e^{-2})$$ $$P(X_{i+1}=III|X_i=I)=P(Y_A>0,Y_B=0)+P(Y_A=0,Y_B=0)=e^{-2}$$ Using identical logic we arrive at a $3\times3$ transition probability matrix delineated below: $$\begin{equation*} P = \begin{bmatrix} (1-e^{-6})(1-e^{-2}) & e^{-6}(1-e^{-2}) & e^{-2} \\ (1-e^{-5})(1-e^{-5/3}) & e^{-5}(1-e^{-5/3}) & e^{-5/3} \\ (1-e^{-4})(1-e^{-4/3}) & e^{-4}(1-e^{-4/3}) & e^{-4/3} \end{bmatrix} \end{equation*}$$ I think I am in good standing with part (a), but part (b) is still confusing. Would solving part (b) require us to find the steady$-$ state probabilities which would effectively reveal the the proportion of the time the shuttle is in state $I,II$ and $III$? Thank you for any insight.