= Solution
Conservation of total <probability> gives $A(t)=1-B(t)$. The two-state <master equation> is therefore
$$
\dot B=k_f(1-B)-k_bB=k_f-(k_f+k_b)B,\qquad B(0)=0.
$$
Solving this linear <ordinary differential equation> gives
$$
\boxed{B(t)=\frac{k_f}{k_f+k_b}\left(1-e^{-(k_f+k_b)t}\right).}
$$
The middle panel in <2/d> sketches this solution for the explicitly illustrative ratio $k_f/(k_f+k_b)=0.6$. The curve starts at zero with slope $k_f$, is increasing and concave down, and approaches the horizontal limit $B_\infty=k_f/(k_f+k_b)$ on <time constant> $(k_f+k_b)^{-1}$. Its initial tangent meets the asymptote at that <time constant>. The complementary limit is $A_\infty=k_b/(k_f+k_b)$, so $B_\infty/A_\infty=k_f/k_b=e^{-\Delta G_{AB}/(k_BT)}$, the <Boltzmann distribution> ratio. These are occupancy <probabilities>; $B(t)$ is not the cumulative first-passage probability when reverse transitions remain possible.
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