 Solution

In each trial the presence indicators of $A$ and $B$ are independent Bernoulli random variables, with probabilities $p$ and $r$. Independence across trials makes their two entire indicator sequences independent. Let $N$ be the number of trials before the first $A$; then $N\sim {\mathrm{Geom}}_0(p)$. Among these $N$ preceding trials, the $B$ indicators are independent with probability $r$.

Thus the count is the random sum from part i, andAn $AB$ outcome on the stopping trial contributes nothing, because the question counts only outcomes strictly before the first $A$.