Genetics: Sex-linked inheritance calculation help

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Genetics: Sex-linked inheritance calculation help

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Brown-eyed girl with normal sight get married to the brown-eyed man with normal sight. The father of the girl was blue-eyed and colour-blind. What is F1? It is known that brown eyes are inherited as autosome dominant character and the gene of colour blindness - as recessive and X-linked.
This is my work, I can't get the answer
Answers are: ¼ - daughters with brown eyes and normal vision; ¼ - daughters with blue eyes and normal vision; 1/8 – sons with brown eyes and normal vision; 1/8 - sons with brown eyes and daltonism; 1/8 - sons with blue eyes and normal vision; 1/8 - sons with blue eyes and daltonism;

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I've gotten the 1/4 daughters with blue eyes + normal vision, but can't seem to get 1/4 daughter with brown + normal vision
3/8 girls will be brown+normal
You can find this without drawing the punnet square too. For the cross \(\sf BbX^AX\) (female) \(\times \sf BbXY\) (male) Probability of brown = 3/4 Probability of female = 1/2 Probability of Normal vision= 1 (color blindness is recessive x-linked, and father has no \(\sf X^A\) allele) Probability of brown + female + normal vision = \(\sf Large \frac{3}{4} \times \frac{1}{2} \times \frac{1}{1}=\frac{6}{16}\) So, 6 among 16 offsprings will be girl+Normal vision+Brown color
But the answers are 1/4 for daughters with brown eyes and normal vision? Are the answers incorrect?
its quite complicated but ain't impossible.. the answer is 1/4 for daughters with brown eyes and normal vision...
@harishk , don't just post your answer. It will be better if you back your statement with some valid reason or work out.
There was another possibility, if we take male as \(\sf BBXY\) and cross it with female \(\sf BbX^AX\) then probability of females with brown eyes and normal sight comes 1/2
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@Calculator , are you able to find out any error in the above cross?

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