Cystic fibrosis in an autosomal recessive disease in humans

Cystic fibrosis in an autosomal recessive disease in humans. Mr. and Mrs. Doe are both heterozygous for a low penetrance disease allele, CFTR Arg117His. The estimated penetrance of the disease is 10%. What is the likelihood that their first child will not be affected?

The answer is 97.5% but I do not understand how. Please show me!

Solution

The penetrance of a disease-causing mutation is the proportion of individuals with the mutation who exhibit clinical symptoms. For example, if a mutation in the gene responsible for a particular autosomal dominant disorder has 95% penetrance, then 95% of those with the mutation will develop the disease, while 5% will not.

Here in the above question, it says 10% penetrance for cystic fibrosis. That means only only 10% of the children with the genotype will have the phenotype expressed i.e they suffer from disease. which also means that 90% of the time the child is unaffected. Since the disease is an autosomal recessive type, only 1 in 4 children will have that genotype ie 25%. That means 75% chance that the first child will not be affected..

Due to low penetrance 90% wont be affected and due to independent assortment 75% wont get those homozygous recessive genes