Hardy-Weinberg Color-blind In a city, 4% of male population have color blindness. How many of the female are (a) color blind carrier, (b) color blind? Suppose the city holds Hardy Weinberg equilibrium. My progress: 4% of male have color blind => $p=F(cb~allele)=0.04$ and therefore $q=F(not~cb)=0.96$. Since HW equilibrium stand, we get the allele frequency among female is the same as among male. Then (a) $2pq=2*0.04*0.96=0.0768=7.68\%$ and (b) $q^2=0.9216=92.16\%$. Am I correct?

A: wild-type allele / a: color blind allele

Because color blindness is recessive and X-linked your assumption $p=F(a)=4\%$ is correct as men do only have one copy of the allele. Subsequently $F(A)=q=1-p=0.96$ is also correct. Therefore:

a) $F(Aa)=2pq=7.68\%$ is correct and b) is wrong, a is the color blind allele and $F(a)=0.04$ therefore it's $p^2=0.04^2=0.0016=0.16\%$.

92% color blind among females seems a bit high :).