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Probability confusion

This is probably a very stupid question.

I was trying to answer a genetics question:

Basically, there is a woman who does not have cystic fibrosis (let's say caused by the recessive allele, a).
That means she is either homozygous dominant (AA) or heterozygous (Aa).

We from the question that her parents are both heterozygous (Aa).

I'm trying to find the probabilty of the woman being a carrier of the recessive allele (heterozygous).

I can draw a table of the possibilities of her parents' offspring:

. .A. .a
A AA Aa
a Aa aa

Now, here is my stupid conundrum. We know she can't be aa, so does that mean we exclude the aa part? So the probability of her being heterozygous is 2/3?

Or is it still 1/2?

Reply 1

Czernek1

so it would sort of be both? you do exclude the aa. But the only 2 possibilities for her are either AA or Aa, you dont need to draw out the punnet square. So it's still 50% I believe? correct me if I'm wrong tho

Reply 2

username5106904

Original post by Czernek1

But the Punnet square shows the relative probabilities of each genotype - there is not an equal chance of AA or Aa, as Aa will be produced twice as often as AA in this scenario, so surely it cannot be 0.5 for that reason

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