Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 45 / 3 / d / Solution

Let $E_q$ be the event that all eight selected subjects carry two copies of the same specified ancestral copy $q$, and that no other generation-4 subject is autozygous at this genetic locus. The selected subjects occur in every sibship, so this event forces both generation-2 transmissions and all six relevant generation-3 transmissions. Applying the preceding conditional probabilities,Once all six parents carry $q$, their other copies come from the two unrelated outside pedigree founders, one on each side of the pedigree. A child not receiving two $q$ copies therefore cannot become autozygous for a different founding copy. This verifies the “only” condition, rather than merely excluding homozygosity for $q$.

For the event $E$ without specifying which of the original four copies is shared, the four alternatives $E_q$ are disjoint: a selected child cannot simultaneously have both its copies descended from two different ancestral copies. HenceAll selected individuals are then autozygous for the same copy and mutually 2-IBD. The distinction from the $3/4$ probability in the alternative interpretation of part (a) is precisely this disjointness.