a) The unaffected grandparents have sibling with cystic fibrosis (cc). So, the parents should be heterozygous. For the heterozygous parents, to have unaffected child, the probability is 2/3.
In the case of man, his unaffected parents having the probability of 1/2 to be carriers of the recessive allele. The probability of the man receive that recessive allele from his parent is 1/2.
So, the probability of the man to be a carrier of the recessive allele = 2/3 ×1/2× 1/2 = 1/6.
The same is with the woman also.
The woman to be carrier of the recessive gene is 2/3 × 1/2 × 1/2 = 1/6.
The probability of both giving the recessive allele (c/c) to the child is 1/4.
So, the probability that the child will have cystic fibrosis = 1/6 ×1/6 × 1/4 = 1/144.
b) If both the parents are heterozygous, there is 3/4 chance, for a normal child and 1/4 chance, for a cystic fibrosis child. Each progeny's genotype is independent of the other.
The child with cystic fibrosis may the first child or second child or third or fourth. If we consider the affected child to be first born, the probability of getting the specified outcome is 1/4 × 3/4 × 3/4 × 3/4 = 27/256.
The probability of getting 3:1 ratio = probability, for four possible birth orders X probability of both parents to be heterozygous.
= 4× 1/6 × 1/6 × 27/256 = 3/256
c) The first child has cystic fibrosis.
The parents are heterozygous.
There is 3/4 probability, for any future child to be normal.
As, each progeny is independent of the other child, the chance of next three children to be normal is 3/4 × 3/4 × 3/4 = 27/64.