Arrivals of buses at a bus stop follow a Poisson process of arrival rate $\lambda$ > 0. Each bus has a probability p of being full when arriving at the stop, independent of other buses. A traveller at the bus stop wishes to catch the next non-full bus, provided it arrives within t units of time. He decides to catch a taxi once the period of t time units expires. Assume throughout that nobody gets off any bus at the bus stop.
(a) How can I show that the numbers of full and non-full buses arriving within the period of t time units are independent Poisson random variables? If so what will the parameters of their respective Poisson distributions be?
(b) Given that a total of m full buses arrive at the bus stop within the period of t time units, how do you calculate the probability that the traveller has to catch a taxi?
a) A Poisson random variable can be expressed as a sum of Exponential random variables, in fact the Poisson law $P(n)$ is the convolution of $n$ exponentials. Now the probability that $n_1$ full buses and $n_2$ non full buses arrive in $[0,t]$ is given by:
\begin{equation} \tilde P(n_1,n_2) = \frac{\lambda^{n_1+n_2}}{(n_1+n_2)!}e^{-\lambda}\binom{n_1+n_2}{n_1}p^{n_1}(1-p)^{n_2} = \frac{(p\lambda)^{n_1}}{(n_1)!}e^{-p\lambda} \;\frac{((1-p)\lambda)^{n_2}}{(n_2)!}e^{-(1-p)\lambda} \end{equation}
Where the binomial coefficient in the first expression counts the number of ways you can have $n_1$ full buses and $n_2$ non full buses. So you see that the joint probability factorizes: $\tilde P(n_1,n_2)= P(n_1)P(n_2)$. This automatically means that the two random variables "number of full buses arrived in $[0,t]$" and "number of non-full buses arrived in $[0,t]$" are independent random variable. We furthermore see that they are Poisson random variables respectively with mean $p\lambda$ and $(1-p)\lambda$.
b) We are asked to compute the probability that $0$ non-full buses arrive given that $m$ full buses arrive. Since the number of full and non full buses are independent, we have: \begin{equation} Prob(n_1=0|n_2=m) = Prob(n_1=0) = P(0) = e^{-(1-p)\lambda} \end{equation}