If a female that is a carrier for color-blindness has a child with a male that has normal vision, what percentage of their children may be color blind?

0%
25%
50%
75%

Color blindness is a sex-linked trait, meaning it is carried on the X chromosome. A female's genotype is XX, whereas a male's genotype is XY. Since the trait is recessive, if one of the X chromosome's on a woman carries the colorblindness but the other chromosome does not, the other chromosome is dominate over the colorblind chromosome and she is a carrier of colorblindness, but not colorblind. 

If a man's X chromosome carries colorblindness, he has no other X chromosome to dominate over the trait, so he WILL be a carrier and will be colorblind. 

Male children of a carrier mother and non-colorblind father have a 50% probability of being colorblind and female children have a 50% probability of being carriers, but the daughters will not be colorblind. This is because one X comes from the mother (which may or may not have the colorblind trait), and one from the father. Even if the mother passes on colorblindness, the father cannot, so the X that comes from the father will dominate over the colorblind trait. 

Male children of a father with colorblindness and a carrier mother have a 50% likelihood of being colorblind, and their female children are either carriers or colorblind. 

Male children of a father with colorblindness and a non-colorblind mother will not inherit colorblindness, but all of the female children will be carriers. 

Male children of a non-colorblind father and a colorblind mother will be colorblind and all the female children will be carriers. 

All the children of two colorblind parents will be colorblind.To recap, in order for a male to be colorblind, the X passed on from his father must contain the colorblind gene. In order for a female to be colorblind, both the X from her mother and the X from her father must contain the colorblind gene, meaning her mother is either colorblind or a carrier and her father is colorblind. We can exhibit these probabilities in a punnet square. Assume that X is a non-colorblind gene, x is colorblind, and Y is non-colorblind. 

If a mother is a carrier, her genotype is Xx. If the father is not colorblind, he is XY. Here is the Punnet square (I doubt this will space properly, but I hope you get the idea): 

| X | x 
 
X | XX | Xx 
Y | XY | xY 

The mother is the first horizontal line, Xx. She is a carrier. The first vertical line is XY, the father, who is not a carrier. You multiply the first mother's gene (X) with the first father's (X) to get the top-left box (XX). The second mother's gene (x) with the father's first gene (X) yields the top, right box (Xx), etc. This is the probability of having a colorblind child. We can see it's true that's there's a 50% chance of having a boy (there are two xy's) and a 50% of having a girl (two xx's). If you have a boy, there's a 50% of having a colorblind boy, and a 50% chance of having a non-colorblind boy. There is a 0% of EVER having a colorblind daughter in this situation. Now let us look at the possible answers: 
A) 25% of their offspring will be color-blind 
^^ Statistically, this is true. If they have four children, then statistically they will have two daughters, two sons, and one of the sons will be colorblind. This is not a certainty, however, because it's possible that they have four daughters, none of which are colorblind. 
B) None of their daughters will be color-blind 
^^ This is unequivocally true, there is no possibility of a daughter being colorblind. 
C) Half of their sons will be color-blind. 
^^ Statistically speaking, this is true. Half of their sons would probably be colorblind, but statistics is not a certainty. It is possible to have two sons in a row, and both of them to inherit an XY genotype, meaning neither of them are colorblind. So this statement is false. 
D) Half of their daughters will be carriers of color-blindness. 
^^ Again, this is statistically true, but genetics is never a certainty in situations like this. 

If this is a question where only one answer can be correct, then the answer is B. If multiple answers can be correct, then A, B, C, and D are true. If you refer to the Punnet square I made above, you can see that 25% of the offspring in the square are colorblind, none of the daughters are, half the sons are, and half the daughters are carriers. They are true when you are speaking in terms of probability, which is all genetics really is. So either they are all true, or if you can only pick one, it is B, because that is the only 100% certainty.