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Label each parent's two allele copies by its realized transmission: let $(A,B)$ be the transmitted and untransmitted alleles of the first parent and $(C,D^{\prime })$ those of the second. Under random mating and Hardy-Weinberg equilibrium, the four parental allele draws are independent with population frequencies $\pi$. Random Mendelian segregation merely swaps the two independent copies within each parent, so the transmitted/untransmitted labelled pair still has probability ${\pi }_a{\pi }_b$; the second pair has probability ${\pi }_c{\pi }_d$.

The child receives $A,C$. If $\mathcal{D}$ is its disease event, its penetrance is $k{\psi }_A{\psi }_C$ and its overall disease probability is $k{Z}^2$. ThereforeThis proves the transmitted and untransmitted alleles under multiplicative penetrance factorization. Under these assumptions, the transmitted pair $a/c$ has the affected-case distribution, while the complementary pair $b/d$ has the population distribution and is independent of the transmitted pair. It is the family pseudo-control genotype.

As in part (a), the formula records labelled transmissions. When observed genotypes are unordered, sum over every compatible parental-origin transmission. This includes multiplicities for overlapping parental alleles or two heterozygous parents producing a heterozygous child; it avoids interpreting one product as the total probability of all observationally identical configurations. The homogeneous random-mating model is also essential to the unconditional independence: mixing ancestry strata can correlate the case and pseudo-control through their shared family background.