To determine the chances that the daughter would be color blind, we need to understand the genetics of color blindness. Color blindness is most commonly caused by defects in the X chromosome, and it is a recessive trait. Since males (XY) have only one X chromosome and females (XX) have two, a male will be color blind if his single X chromosome carries the defect. A female, on the other hand, would need to have the defect on both of her X chromosomes to express color blindness since it is recessive.

Given that the man has normal vision, we can infer that his X chromosome (which he got from his mother) does not have the color blindness defect; however, because his father was color blind, we know that he carries a Y chromosome without the color blindness trait (otherwise, the man would be color blind too). The man will pass on either his X or his Y chromosome to his offspring. If it's a daughter, he will pass on the X chromosome, which we know does not carry the color blindness defect.

The woman's father was color blind, meaning her father's only X chromosome carried the defect. Since women have two X chromosomes, the one she received from her mother could potentially not have the defect. Therefore, the woman can be heterozygous (one normal X chromosome and one with the color blindness defect) or homozygous normal (both X chromosomes without the defect). However, the problem does not provide information about the mother's vision or genotype to confirm whether she is a carrier or not. If the woman is not a carrier, none of her children would inherit color blindness. If she is a carrier, then there's a 50% chance she could pass on the X chromosome with the defect.

So, let's consider the two potential scenarios for the woman, represented by X (normal X chromosome) and X^c (X chromosome with color blindness defect):

1. The woman is a carrier (X X^c): There's a 50% chance she might pass on the X chromosome with the defect (X^c) since she is heterozygous.

2. The woman is not a carrier (X X): There's a 0% chance the child will be color blind since all her X chromosomes are without the color blindness defect.

In conclusion:

• If the woman is a carrier, the daughter has a 50% chance of getting the X^c chromosome from her mother and would then be a carrier like her mother (heterozygous), but will not be color blind because the X chromosome from her father is normal.
• If the woman is not a carrier, then the daughter has a 0% chance of being color blind because both of her parents would provide normal X chromosomes.

Because we don't know the mother's carrier status with the information given, we cannot say for certain what the probability is for the daughter to be color blind. However, we know it's either 0% (if the mother is not a carrier) or 50% (if the mother is a carrier) chance of being a carrier, but 0% chance of expressing color blindness since she receives one normal X chromosome from her father. Therefore, the probability that the child would be color blind (express the trait) is:

Option D: Zero percent.