Two individuals are heterozygous for a recessive blindness trait. They had a kid. She shows no symptoms...

what is the probability that she is a carrier? Lets say A/a = Not Blind/Blind.

If The mother gave the recessive 'a' (p=1/2), then the father must have given 'A' (P=1). If the mother gave her 'A' (p=1/2) then there was a chance her father gave her 'a' (p=1/2) or 'A' (p=1/2). This corresponds to P=(1/2 x 1) + (1/2 x 1/2) =3/4

Alternatively, when I draw a punnet square, I can cross out the a/a box. Then I can count my heterozygotes, which are now 2/3. Why don't my two methods agree?!

If the parents are both heterozygous, Aa, then Pr(A) form both father and mother = (1/2)(1/2) = 1/4. Pr(A) from father = 1/2 and Pr(a) from mother = 1/2 and (1/2)(1/2) = 1/4. The other way around is also 1/4. So the total probability of "not blind" is 3/4.

Your error is saying that the Pr(father giving "A") = 1, which it is not...

The probability of a blind child, as Asst Prof correctly said, is 1/4 (since 1/2 x 1/2 =1/4). But the question initially asked was if the child can see, what is the probability she is a carrier? You are correct in "crossing out the a/a box" of your Punnet square (as the child can see) to get 2/3.

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