I am dealing with a task from the sample set of exam problems in my course of stochastic processes, I just want to understand whether my approach to the task is correct.
Given:
We wait at a bus stop, shuttle busses and minibuses arriving at it, according to Poisson processes with respective rates of $2$ and $5$ (That is: $S(t)\sim\text{Poisson}(2),~M(t)\sim\text{Poisson}(5)$). Find the probability of at least two minibuses to arrive before a shuttle bus.
My attempt:
Let us denote the time between $m$'s and $(m+1)^\text{st}$ arrival of shuttle busses and minibuses as $\mathbb{S}_{(m,m+1)}$ and $\mathbb{M}_{(m,m+1)}$. We need to find $\mathbb{P}\left(\mathbb{S}_{(0,\geqslant2)} < \mathbb{M}_{(0,1)}\right)$. We can rewrite it as the probability of at most two minibuses to arrive before a shuttle bus: $1-\mathbb{P}\left(\mathbb{S}_{(0,<2)} < \mathbb{M}_{(0,1)}\right)$. Then by formula proposed here, the probability has to be: $1 – \frac{2}{2+5}\cdot\frac{5}{2+5} \approx 0.7959$. However, I am not that confident in the results of mine. One of my colleagues has proposed yet another formula for such a probability: $1-\frac{2}{2 + 2\cdot5} = 0.8\bar{3}$. I have no idea, from where does the second formula come, and I am not that confident in the results of mine.
Can you, please, help me to indentify, whether my solution is correct or not?
Thank you in advance!
Best Answer
Let $S(t)$ and $M(t)$ be independent Poisson processes with parameters $\lambda$ and $\mu$. The probability that there are at least $n$ arrivals from $S(t)$ before the first arrival is given by $$ \left(\frac\lambda{\lambda+\mu}\right)^n. $$ Here $\lambda=2$, $\mu=5$, and $n=2$, for a result of $\frac4{49}$. Your answer of $\frac{25}{49}$ is not assuming independence between the arrival processes