= Solution
Assume identical dwell states and memoryless forward/backward exits with constant rates $k_+$ and $k_-$; put $K=k_++k_-$. The <probability> density for the first exit to be forward at $t_1$ is $k_+e^{-Kt_1}$. For the very next exit also to be forward at time $t$, multiply by $k_+e^{-K(t-t_1)}$ and integrate over the unobserved first time:
$$
\boxed{\Pr(\hbox{next two exits forward, total time in }dt)
=\int_0^t k_+e^{-Kt_1}k_+e^{-K(t-t_1)}dt_1\,dt
=k_+^2t e^{-Kt}\,dt.}
$$
This density integrates to $(k_+/K)^2$, the <probability> that both directions are forward. Conditional on that event, the normalized total-time density is the <Erlang distribution>
$$
\boxed{p(t\mid++)=K^2t e^{-Kt},\qquad t_{\rm mode}=\frac1K\simeq0.010\,\mathrm s.}
$$
Its mean is $2/K$, so the most probable time is not the mean. Since both backward-rate conventions in part (c) have $k_-\ll k_+$, neglecting backwards steps gives the usual approximation $p(t)=k_+^2te^{-k_+t}$ and mode $1/k_+$. If “two forward steps” instead permits intervening backwards events, thinning independent Poisson exits gives that latter rate-$k_+$ distribution directly; it describes a different conditioning. The low barrier noted in part (c) limits the physical accuracy of all these memoryless dwell assumptions.
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