Answer:
27⁄64
Explanation:
A recessive disease is expressed only when an individual carries two copies of the recessive allele (i.e., individuals must be recessive homo-zygous to suffer from the disease). In this case, both parents are carriers for hemochromatosis, i.e., both are heterozygotes carrying the defective allele, thereby the probability of having a child with a normal 'dominant' phenotype is 3/4, i.e., 3/4 individuals are expected to have an A_ genotype (1/2 AA + 1/4 Aa = 3/4), and 1/4 individuals are expected to have an aa genotype (where 'A' is the dominant allele and 'a' is the recessive allele associated with hemochromatosis). In consequence, the probability of having three children with the normal phenotype is 3/4 x 3/4 x 3/4 = 27/64.
