Disease-ascertained transmission probability ($P(g\mid g_f,g_m,D)=\frac{q_g{R}_g}{{\sum }_h{q}_h{R}_h}$)

Let $q_g=P(g\mid g_f,g_m)$ be the Mendelian segregation probability, and assume the child's disease status depends on the parental genotypes only through its own genotype, with penetrance proportional to $R_g$. Bayes' theorem gives the displayed conditional law. For a diallelic parent pair there are four equally likely labelled transmission outcomes. If an unordered genotype occurs in $m_g$ of them, $q_g=m_g/4$, so its probability is $m_g{R}_g/{\sum }_h{m}_h{R}_h$. Dropping multiplicities would, for example, misweight a heterozygous child of two heterozygotes.