In humans, the allele for red-green color blindness is x-linked and recessive (xr; the dominant allele is called xr). a man with normal color vission marries a color-blind woman who gives birth to a color-blind daughter. from this, we can tell that

Allele for red-green colour blindness is x-linked recessive.
The XR allele is a normal (dominant) for red-green colour blindness, while the Xr allele is recessive.

This means that for a male who has only one X chromosome, carrying the XR allele (hemizygous dominant) means he has normal vision, while carrying the Xr allele (hemizygous recessive) will result in red-green colour blindness for him.

The situation is slightly different for a female, who has two X chromosomes.
Since the abnormality is recessive, both X-chomosomes must carry the Xr allele for her to develop red-green blindness (homozygous recessive, or XrXr).  In the other cases homozygous dominant (XRXR) will be carrying both normal genes.  Heterozygous (XRXr) will make the woman a carrier.

If the father has normal vision, then he cannot be a carrier (he has only one X chromosome), so he has the XR allele.
If the mother red-green colour-blind, then she must be a homozygous recessive, i.e. XrXr alleles.

If the offspring is a boy, then he must be red-green colour-blind, because the only genotype is XrY, the Xr from the mother, and Y from the father.

The only possible genotype for a daughter is XRXr, since one of the X-chromosomes must come from the father, and the mother (homozygous recessive, XrXr) provides the defective allele.  So the only phenotype for a daughter is normal vision.

From this, we can tell that it is not possible for a father with normal vision to have a daughter with red-green colour-blindness, contrary to what the question describes.  [ => some genetic testing is in order, perhaps ]

Try and make the question more clear : )

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