{--
Now, in conjugation, each parent gene donates either allele with equal 
likelihood to the child gene.

So, for example, you have two parents, one with genotype Aa, the other
with genotype AA, the child has one of the following four possible genes:

 Aa x AA = (AA, AA, Aa, Aa) or, simplifying: (AA, Aa) each option equally likely

So, today's Haskell problem. Given a genotype, one from each parent, define
the function, cross, that gives a probability distribution of the resulting
genotype in the offspring.
--}

cross :: Genotype -> Genotype -> Prob Genotype
cross a b = undefined
cross :: Genotype -> Genotype -> Prob Genotype
cross a b = undefined
cross a b = undefined