Let the arrivals from Bus A be a Poisson process with rate 2/hr and the arrivals from Bus B be a Poisson process with 7/hr. Both processes are independent.
1) What is the probability that the first Bus A arrives before the first Bus B?
I think I solved this, just wanted to check if my method was right. So far, I calculated this to be $\frac{2}{9}$ given that $P(T_A<T_B)$ =$\int_{u=0}^\infty\int_{v=u}^\infty$$2e^{-2u}7e^{-7u}$=$\frac{2}{9}$.
2) If each bus breaks down independently with probability $\frac{1}{2}$ before reaching the stop, what is the probability that someone waits for 30 minutes without seeing a single bus?
So far, I separated this into 4 cases: 1) Both buses take more than 30 minutes without breaking down, 2) Bus A breaks down, Bus B takes more than 30 minutes without breaking down, 3) similar to scenario 2) but reversed, and 4) both busses break down.
I found the total probability to be: $e^{-2/2}e^{-7/2}$ + $\frac{1}{2}*e^{-7/2}$ + $\frac{1}{2}*e^{-2/2}$ + $\frac{1}{2}$* $\frac{1}{2}$ $\approx$ $0.46$
Is this approach correct?
I agree with the reasoning in (1). For (2), I believe we can thin the Poisson process to obtain a new one $N_t$ with rate $0.5 * 9$, which is the process of buses arriving (i.e. the buses which do not break down). Note that I've merged the separate Poisson processes for each bus into one. Then we are asking for the probability that $N_{0.5} = 0$, which is just given by $$ \frac{(9/4)^{0} * e^{-9/4}}{(0!)} = e^{-9/4} \approx 0.11$$ since $N_{0.5}$ has distribution Poisson with rate $0.5 * (0.5 * 9)$.