 Solution

Let the prior odds be $\mathbb{P}(T)/\mathbb{P}(F)=R$, so $\mathbb{P}(T)=R/(1+R)$. The Type I error gives $\mathbb{P}(P\mid F)=\alpha$, while the statistical power gives $\mathbb{P}(P\mid T)=1-\beta$. By Bayes' theorem,This is a posterior probability of the null, not the false-positive probability $\alpha$ conditional on the null. The distinction is the base-rate effect in a positive study.