 Solution

Within the active set, cross-classify dosage $Z_i\in \{1,2\}$ against $Y_i$. Conditional on the vector $A=(A_1,\dots ,A_n)$, complete randomization assigns $n_1$ of the active patients to low dosage and $n_2$ to high dosage uniformly. Under independent assignment, active patients receive low and high dosage with conditional probabilitiesconditioning further on their low- and high-dose totals again gives the same uniform allocation. Under $H_B$, active patients' outcomes are fixed as their common $Y_i(1)=Y_i(2)$, so the conditional Fisher p-value obeys

Under $H_A$, $P_A$ is a function only of $A$ and the fixed outcomes. Applying the law of iterated expectation under the joint null givesThis conditional argument explains the stated near-independence even though the two tests reuse outcomes.