= Disease-ascertained transmission probability
{title2=$P(g\mid g_f,g_m,D)=\frac{q_gR_g}{\sum_hq_hR_h}$}
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 $m_gR_g/\sum_hm_hR_h$. Dropping multiplicities would, for example, misweight a heterozygous child of two heterozygotes.
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