= Solution
Let $K$ count second-fork arrivals. Its independent increment over $(s,s+t]$ is Poisson$(rt)$, and
$$
\mathbb P(K(s)\text{ even})=\frac{1+e^{-2rs}}2,\qquad
\mathbb P(K(s)\text{ odd})=\frac{1-e^{-2rs}}2,
$$
obtained 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. Therefore
$$
\boxed{\begin{aligned}
\mathbb P(X_A(s+t)=X_A(s))&=e^{-\lambda t/3},\\
\mathbb P(X_B(s+t)=X_B(s))&=e^{-rt}\left[1+\frac{rt}2(1-e^{-2rs})\right],\\
\mathbb P(X_C(s+t)=X_C(s))&=e^{-rt}\left[1+\frac{rt}2(1+e^{-2rs})\right].
\end{aligned}}
$$
At $s=0$ these reduce to the distinct initial-delay survivors; as $s\to\infty$, both non-Poisson probabilities approach the equilibrium-delay survivor from part (b).
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