The female i-1 and her mate, male i-2, had four children, one of whom has albinism. what is the probability that they could have had a total of four children with any other outcome except one child with albinism and three with normal pigmentation?

75% chance of having four normal phenotype children without albinism.

50% chance of having four children who are carriers of the gene with normal phenotype.

Explanation:

Albinism results from an autosomal recessive gene.  In order to solve this problem we have to draw a punnet square and figure out the genotypes of the parents.  So with the genotype of the parents both parents have to be carriers of the trait since it is an autosomal recessive trait.

Genotype of female I-1 = Aa

Genotype of male I-1 = Aa

PARENTSAa

A                AA  Aa

a                Aa  aa

75% chance of having four normal phenotype children without albinism.

50% chance of having four children who are carriers of the gene with normal phenotype.

25% chance of having children with albinism.

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explanation:

you know they are heterozygous for a trait because they have it as jj or jj.