Since colorblindness is a recessive disease, all copies of the X-chromosome must have the diseased allele in order for the person to be colorblind. Daughters have two copies of the X-chromosome: one from the mother and the other from the father. Males only have one copy of the X-chromosome (from the mother) and a Y-chromosome from the father.

Since we know that the father has normal vision, he does NOT carry the colorblind allele. Since the daughters for this couple can only potentially receive one colorblind allele (from the mother), all of their daughters will have normal vision. This means that there is a zero percent chance for colorblindness in their daughters.

The cross would look like this, taking X^b as the colorblind allele:

Parents: XX^b x XY

Offspring: XX or XX^b (normal daughters), XY (normal son), YX^b (colorblind son)

The chance of a colorblind daughter will be zero, but the chance of a colorblind son will be 50%.

X-linked recessive inheritance dictates that expression of themutant phenotype will only occur if the individual is homozygous for the mutation on the X-chromosomes. Therefore, a female must have inherited two mutant X-chromosomes to have hemophilia A, while a male only requires one mutant X-chromosome to have the disorder. By virtue of the father not having hemophilia A, we know the daughter is inheriting at least one wild-type X-chromosome, and therefore there is zero chance she will be homozygous and have hemophilia A.

Incomplete dominance indicates that there is no dominant allele. In these cases, the phenotype associated with inheriting one copy of each allele (the heterozygotes, Bb) is often a blending of the phenotypes associated with homozygosity of each allele. As such, a genotype of BB will result in black hair, bb will produce white hair, and Bb will result in grey hair.

The incorrect answers are too limited in scope to be cases of incomplete dominance. The correct answer identifies that there will be three unique phenotypes. 

Because the disorder is autosomal dominant, the statement "If an individual does not have the disorder, they can still pass on the mutant gene if one of their parents has the disorder" must be false.

If the indivdual in question does not have the disorder, that means they did not inherit ANY copies of the mutant gene, and therefore cannot pass it on.

For autosomal dominant disorders, the individual only needs to inherent a single copy of a mutated allele to then show symptoms of that disorder. If it were recessive, both alleles would have to be mutant. X-linked recessive is incompletely correct for males since they only have one X-chromosome, and incorrect for females since 2 copies of the X-chromosome are needed, and thus 2 copies of the allele. Complex inheritance describes situations beyond a single gene, and Mendelian inheritance is not a specific method of inheritance. Note that Y-linked disorders are passed from father to son, and since males only have one copy of the Y-chromosome, if there is a genetic mutation on the Y-chromosome, the individual will be affected.