= Solution
{c}
For an interior true parameter and large $n$, standard <asymptotic normality of a maximum likelihood estimator> gives
$$
\boxed{\widehat p\ \dot\sim\ N\left(p,\frac{p^2(1-p)}{nr}\right).}
$$
Estimate its standard error by $s_e=\widehat p\sqrt{(1-\widehat p)/(nr)}$. A two-sided approximate <Wald confidence interval> is $(L,U)=(\widehat p-z_{1-\alpha/2}s_e,\widehat p+z_{1-\alpha/2}s_e)$, intersected with the parameter range $(0,1)$. The normal quantile is chosen so a standard normal lies between its positive and negative values with probability $1-\alpha$.
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