= 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>,
$$
P(E_q)=\frac14\frac1{64}\left(\frac14\right)^8\left(\frac34\right)^8
=\frac{3^8}{2^{40}}.
$$
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. Hence
$$
\boxed{P(E)=4P(E_q)=\frac{3^8}{2^{38}}.}
$$
All 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.
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