Genetic problem please help i have a test tomorrow and i think am gonna fail :(?

tell me how u would do this problem:
Color blindness is a sex linked trait and curly hair is dominant over straight hair. A man who is color blind and is heterozygous for curly hair marries a women who is a carrier for color blindness and has straight hair. List the phenotypic ratio of their children. you should have a: color blind b: colorblind curly hair c: normal vision curly d: normal vision straight. PLEASE if u know it try to explain it to me please please please i really want to pass that test on thursday! thnx plus for the best answer 10 points!

Color blindness is a recessive sex linked disorder. so the genotype for that is XcXc for a female and XcY for a male. If they have normal vision, the female can be either homozygous X+X+ (the pluses symbolize that lack of the mutant trait) or heterozygous X+Xc since the colorblindness allele is recessive, the heterozygous phenotype is normal vision. For curly hair, I'm almost positive that is autosomal. So that is the normal CC, Cc, cc being Homozygous curly, heterozygous curly, and straight respectively. Therefore you must perform a punnet square with 2 traits being evaluated. So you man who is colorblind is for sure XcY since he must have at least one colorblindness allele, and he is Cc for hair. The wife is X+Xc since she is a carrier of the allele, and she is cc for hair. I believe you can do the punnet square from there right? You should get a 9:3:3:1 ratio of offspring in F1generation. But this one is different not a 9:3:3:1 model like usual.
your mom has only 2 possible gametes to arrange across the top of you punnet square, and those are X+ c or Xc c.
The male however has 4 possible gametes to line up down the side of your punnet square, and those are Xc C, Xc c, Y C, and Y c. This is assuming the 2 alleles are not linked of course! I don't think these two genes are linked or if they are epistatic, polygenetic or anything, but assuming normal independent assortment, that should give you a 2 by 4 punnet square yielding 8 different genotypes. From the results I got you have XcX+ Cc, XcX+ cc, X+Y Cc, X+Y cc, XcXc Cc, XcXc cc, XcY Cc, XcY cc. those are your 8 genotypes for F1.

SO.... here is your answer...you have respective to my genotype list above: normal/ curly, normal/ straight, normal/ curly, normal/ straight, colorblind/ curly, colorblind/ straight, colorblind/curly, colorblind/ straight. That is
2 normal/ curly
2 normal/ straigth
2 colorblind/ curly
2 colorblind/ straigth

ok. there would be a 50% chance of colorblindness. and a 75% chance of straight hair.