= Solution
The <disease-ascertained transmission probability> follows directly from <Bayes' theorem>. Assume that, given the child's genotype $g$, its disease probability is a common baseline <penetrance> times the <genotype relative risk> $R_g$, with no additional dependence on the parental genotypes. Write $q_g=P(g\mid g_f,g_m)$ for the <Mendelian segregation> probability. Then
$$
\boxed{P(g\mid g_f,g_m,D)=\frac{q_gR_g}{\sum_hq_hR_h}.}
$$
There are four equally likely labelled parental transmission outcomes. For one specified outcome $t$, this becomes $P(t\mid g_f,g_m,D)=R_{g(t)}/\sum_{t'}R_{g(t')}$, which is the four-outcome interpretation of the displayed formula.
For an unordered child <genotype>, let $m_g$ be the number of those outcomes producing it. Its probability is instead
$$
\boxed{P(g\mid g_f,g_m,D)=\frac{m_gR_g}{\sum_hm_hR_h}.}
$$
For example, two heterozygous parents have multiplicities $(1,2,1)$ for $11,12,22$. A heterozygous child therefore has probability $2R_{12}/(R_{11}+2R_{12}+R_{22})$. The four possible outcomes must be counted with multiplicity; treating their unordered genotypes as four distinct equally likely values would be incorrect.
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