 (What if \(p(E) = 1\), though? Then \(H\) may fit \(E\), but so may \(\neg H\). If \(p(E)=1\), we can prove that \(p(E\mid H)=1\) and \(p(E\mid \neg H)=1\) (hint: combine The Law of Total Probability with Complementarity for Contradictories). In other words, \(E\) fits both \(H\) and its negation perfectly. So it shouldn’t be able to discriminate between these two hypotheses. And, indeed, in this case \(p(H\mid E)\) comes out the same as \(p(H)\), so \(c(H,E)=0\).)