 Solution

Let $K$ count second-fork arrivals. Its independent increment over $(s,s+t]$ is Poisson$(rt)$, andobtained by evaluating $\mathbb{E}(-1{)}^{K(s)}=e^{-2rs}$. If $K(s)$ is even, the next second-fork arrival goes to B, so no B arrival requires zero new arrivals. If it is odd, the next goes to C, so up to one new arrival is allowed. Reverse these alternatives for C. ThereforeAt $s=0$ these reduce to the distinct initial-delay survivors; as $s\to \infty$, both non-Poisson probabilities approach the equilibrium-delay survivor from part (b).