1. **Stating the problem:** We are given that red-green color blindness is a sex-linked recessive trait. A color blind woman marries a man with normal color vision. We need to find the probability that their daughter will be color blind. 2. **Understanding the genetics:** Color blindness is a recessive trait linked to the X chromosome. Females have two X chromosomes (XX), males have one X and one Y (XY). 3. **Genotypes:** - The woman is color blind, so her genotype must be X^cX^c (both X chromosomes carry the recessive color blindness allele). - The man has normal vision, so his genotype is X^CY (X chromosome with normal allele, Y chromosome). 4. **Possible gametes:** - Woman can only pass X^c. - Man can pass X^C or Y. 5. **Offspring genotypes:** - Daughters get one X from each parent: X^c from mother and X^C from father, so daughters are X^CX^c (carriers, but not color blind). - Sons get X^c from mother and Y from father: X^cY (color blind). 6. **Probability daughter is color blind:** Since daughters get X^C from father and X^c from mother, they will be carriers but not color blind. **Final answer:** The chance their daughter will be color blind is 0 percent. $$\boxed{0\%}$$