In the given genotypes, "XRX" represents a normal vision allele on the X chromosome, and "Xr" represents a color-deficient allele on the X chromosome. "XRY" represents a normal vision allele on the X chromosome and a normal allele on the Y chromosome.
Since color-deficient vision is a sex-linked recessive trait, it is carried on the X chromosome. In order for a child to have color-deficient vision, they would need to inherit the color-deficient allele (Xr) from their mother and the normal allele (XRY) from their father.
The probability of the child being a carrier for color-deficient vision (heterozygous) is 0.5, and the probability of the child having color-deficient vision (homozygous) is 0.25.
Therefore, the probability that the child will have color-deficient vision is:
P(homozygous) + P(heterozygous) = 0.25 + 0.5 = 0.75
Therefore, the correct answer is B. 0.75.