= Solution
Assume the first-fork choices are independent Bernoulli routings, as in <Poisson thinning>. Put $r=2\lambda/3$. Cars assigned to A form a <Poisson process> of rate $\lambda/3$, also an ordinary <renewal process> with independent exponential holding times of rate $\lambda/3$. The stream to the second fork is an independent <Poisson process> of rate $r$.
B receives its odd-numbered arrivals. Its initial delay is exponential of rate $r$, followed by independent <Erlang distributions> of shape two and rate $r$ between successive B arrivals. Thus \b[B is a delayed <renewal process>], not Poisson. C receives the even-numbered arrivals: its first and all subsequent holding times are independent Erlang$(2,r)$, so \b[C is an ordinary <renewal process>], not Poisson. Their holding-time density and survivor are $r^2t e^{-rt}$ and $e^{-rt}(1+rt)$. Both have asymptotic rate $r/2=\lambda/3$; equality of these mean rates with A does not make the processes Poisson.
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