Relation between Poisson process and geometric distribution 
Suppose that red cars arrive at an intersection according to a Poisson process with rate parameter $r>0$ and blue cars arrive, independently of red cars, according to a Poisson process with rate parameter $b>0$. If $X$ is the number of blue cars that arrive between two successive red cars, show that $X$ has a geometric distribution.

Can anyone help me with this problem? I have tried to rephrase $X$ as: how many cars will arrive at the intersection before we see any red car. 
 A: Let the arrival process of red cars is according to the Poisson process, say, $\{R(t),t\ge 0\}$ with an intensity parameter $\lambda_1>0$ and the arrival process of blue cars is according to the Poisson process, say, $\{B(t),t\ge 0\}$ with an intensity parameter $\lambda_2>0$. Further, both the processes are independent of each other. We are interested in the number of arrivals of blue cars between two successive arrivals of red cars. 
We know that, the inter-arrival times of a Poisson process with parameter $\lambda$ are exponential with mean $1/\lambda$. This means that, the pdf of inter arrival times between successive red cars is $$f(t)=\lambda_1 e^{-\lambda_1 t}.$$ Now, the probability distribution of $X$, the number of blue cars during an arbitrary interval (as determined by successive arrivals of red cars) is
\begin{eqnarray*}
P\{X=k\}&=&\int_{0}^{\infty}\dfrac{e^{-\lambda_2 t}(\lambda_{2}t)^{k}}{k!} f(t)dt\\
&=&\int_{0}^{\infty}\dfrac{e^{-\lambda_2 t}(\lambda_{2}t)^{k}}{k!} \left(\lambda_1 e^{-\lambda_1 t}\right) dt\\
&=&\dfrac{\lambda_1 \lambda_{2}^{k}}{k!}\int_{0}^{\infty}e^{-(\lambda_1 +\lambda_2)t}\cdot t^{(k+1)-1}dt\\
&=&\dfrac{\lambda_1 \lambda_{2}^{k}}{k!} \cdot \dfrac{\Gamma (k+1)}{(\lambda_1 +\lambda_2)^{k+1}}\\
P\{X=k\}&=&\left(\dfrac{\lambda_1}{\lambda_1 +\lambda_2}\right)\left(\dfrac{\lambda_2}{\lambda_1 +\lambda_2}\right)^{k},\quad k=0,1,2,\cdots
\end{eqnarray*}
which is a geometric distribution with parameter $\dfrac{\lambda_1}{\lambda_1 +\lambda_2}$. 
