If a haemophilic man marries a normal but carrier woman, what is the probability of their son becoming haemophilic?

The correct option is A 50%
Haemophilia is an X-linked recessive disorder. Since males are hemizygous for chromosome, one copy of the affected gene in males in each cell is sufficient to cause the disorder (X${}^h$Y or X${}^h$Y ). Females with two copies of the affected gene show the disorder (X${}^h$X${}^h$ or X${}^h$X${}^h$). Females heterozygous for this trait will be normal but serve as a carrier of the disease. Since father transmits its X chromosome to the daughters, not to son, so the haemophilic father would not affect the disease inheritance of his son. Son receives its X chromosome from mother which is carrier (X${}^h$X) which means half of her sons will get normal copy of X chromosome and another half will get the affected copy. Thus, there is 50% probability of their son to have the disease. The only haemophilic mother can have all of her sons affected with the disease but the presence of two copies of affected alleles makes the survival of haemophilic female very low. A carrier mother cannot have all normal sons due to the presence of one copy of affected allele. Thus, the correct answer is option C.