= Solution
Treat the two parents as unordered, so three possible <genotypes> give six mating types. A homozygous parent transmits its allele with probability one; a heterozygous parent transmits either <allele> with probability one half by <Mendelian segregation>. Independent parental transmissions give the following complete offspring distributions:
$$
\begin{array}{c|ccc}
\text{parental genotypes}&\Pr(a/a)&\Pr(a/b)&\Pr(b/b)\\\hline
a/a\times a/a&1&0&0\\
a/a\times a/b&1/2&1/2&0\\
a/a\times b/b&0&1&0\\
a/b\times a/b&1/4&1/2&1/4\\
a/b\times b/b&0&1/2&1/2\\
b/b\times b/b&0&0&1
\end{array}
$$
Each row sums to one. In particular, heterozygote-by-heterozygote mating can produce all three offspring <genotypes>, whereas the two opposite homozygotes produce only heterozygotes. These conditional offspring probabilities do not require <Hardy-Weinberg equilibrium> of the parents.
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