Probability of a passenger going to a city

Question

Buses headed to a city $A$ arrive in a city $B$ every $30$ minutes starting at 8:37am, whereas buses heading to city $C$ arrive $15$ minutes starting at 8:31am.
$1)$ If the passenger arrives at the station uniformly between 8:30am and 9:30am and then gets onto the first bus that arrives, what is the probability he goes to the city $A$?
$2)$ What is the probability if the passenger arrives uniformly between 8:45am and 9:45am?

I am totally confused with this problem. Please help me to solve this.

Answer 1

1) The passenger goes to the $A$ city if he comes to the station in $(8:31am, 8:37am] \cup (9:01am, 9:07am]$. So by the law of the total probability, the probability of going to the city A is $\frac{6}{60} + \frac{6}{60} = \frac{1}{5}$

Answer 2

At whichever time the pasenger arrives, he/she will take the first bus that departs. Lets assume he/she arrives at $t_a$ Thus the bus he/she will take will go at $$t_d = min(min_{t_{A,i} \geq t_a}(t_{A,i}),min_{t_{C,i}\geq t_a}(t_{C,i})),$$where $t_{X,i}$ are the departure times of a bus going towards city $X \in \lbrace A,B \rbrace$. This means, if the passenger arrives between 8:30 and 8:31, he/she will take the ride to $C$. If he/she arrives between at 8:31 and 8:37 he/she will go to $A$. If he/she arrives between 8:37 and 9:07, then $C$ is the destination again ... This way the one hour, in which the passenger may arrive, is divided into intervalls, which correspond to going to $A$ or $C$, respectively. Knowing the length of all the intervalls corresponding to a destination will give you the probability for going there, since you know the total length of the time-window, in which the pasenger arrives with uniform probability...