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?