The first step in solving this problem requires an examination of the possible genotypes of the parents. As stated color-blindness is a recessive sex linked trait. So the genotypes of the parents should be:

mother's genotype(normal vision but whose father was color blind) X^{N}X^{n}She is a carrier because of her heterozygous condition. For the recessive gene to be expressed one must be homozygous for it.

father's genotype(color blind) X^{n}Y

From the Punnett square it is obvious that half of the children will be color blind and the other half will be normal. More specifically there will be1 normal boy to1 color-blind boy to 1 normal girl to 1 color- blind girl.

X^{N}X^{n}one normal (carrier) girlX

^{N}Y one normal boyX

^{n}X^{n}one color blind girlX

^{n}Y one color blind boy

- Probability of having a color blind daughter is (1/2 x 1/2) or 1/4 or 25% (0.25) This is because both sex and colorblindness are involved in the solution.
- The probability that their first son is color blind is 50% (0.50). In this question only males are part of the solution.