= Solution
Use states $0,A,B$, where $0$ denotes an empty molecule and the other states denote its single occupied site. The source arrows give $0\to A$ at rate $f_1$, $A\to0$ at rate $f_4$, $A\to B$ at rate $f_2$, and $B\to A$ at rate $f_3$. Thus
$$
\boxed{\frac d{dt}\begin{pmatrix}p_0\\p_A\\p_B\end{pmatrix}
=\begin{pmatrix}
-f_1&f_4&0\\
f_1&-(f_4+f_2)&f_3\\
0&f_2&-f_3
\end{pmatrix}\begin{pmatrix}p_0\\p_A\\p_B\end{pmatrix}.}
$$
Each diagonal entry is the negative total escape rate from its column's state. In particular, the empty state cannot make a direct jump to $B$: the reservoir touches $A$ alone. Conversely, $A\to B$ is not a second charge arriving on the molecule; it is an internal hop of its existing charge. Summing the three equations gives $d(p_0+p_A+p_B)/dt=0$, which checks the <master equation> convention and the single-charge constraint.
\Image[/past-exam-of-the-theoretical-physics-major-option-course-of-the-university-of-oxford/2015-three-state-charge-transition-graph.png]
{title=Three-state charge kinetics and the two distinct stationary jump counts}
{description=Original state-space graph inferred from the source rates. Each arrow is a charge jump; an A-to-B jump changes location without changing the molecule's total charge.}
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