= Solution
Write the <probability> column vector as $\mathbf p=(p_0,p_A,p_B)^T$, with $0$ denoting the uncharged state. The arrows in the original figure give transitions $0\to A$ at rate $f_1$, $A\to B$ at rate $f_2$, $B\to0$ at rate $f_3$, and $B\to A$ at rate $f_4$. Exclusion of double occupation means injection occurs only from state $0$. Thus the <master equation> is
$$
\boxed{\frac{d}{dt}\begin{pmatrix}p_0\\p_A\\p_B\end{pmatrix}=\begin{pmatrix}-f_1&0&f_3\\f_1&-f_2&f_4\\0&f_2&-(f_3+f_4)\end{pmatrix}\begin{pmatrix}p_0\\p_A\\p_B\end{pmatrix}.}
$$
For example, $\dot p_B=f_2p_A-(f_3+f_4)p_B$. Every allowed jump adds to one <probability> and subtracts from another, which verifies the zero column sums.
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