A sickness has a heterozygote frequency of $\frac{1}{20}$ ie. 1 in 20 people of a population will have the allele combination $Aa$ where a denotes the recessive and A the dominant allele. To become sick the carrier needs to have two $aa$ alleles.
A mother is known to have the combination $Aa$ , what is the probability her child will be sick (have $aa$) if the combination of the husband is unknown (i.e. $AA, Aa, aa)$?
Now here is where I struggle. If the husband has $AA$ , the child will never get $aa$, if the husband has $Aa$, the child will get $aa$ with a chance of $\frac{1}{4}$.
If the husband has $aa$ (is sick himself) then the child wil get $aa$ with a chance of $\frac{1}{2}$
The book states that the answer a priori is for the child to be sick in this situation is $\frac{1}{20} \frac{1}{2} \frac{1}{2} = \frac{1}{80}$
Shouldn't it instead be : $\frac{1}{4}\frac{1}{20} + (\textbf{double aa frequency})\cdot \frac{1}{2}$?
You have mentioned that the father could be AA, Aa, or aa. What we need to remember is that these represent the alleles he received from his parents. So, if we let the first letter represent the allele received from his father and the second letter represent the allele gained from his mother, there are actually four possibilities: AA, Aa, aA, aa. That is, the father has a 1/4 chance of being either AA or aa, but a 2/4 = 1/2 chance of being Aa.
When these are combined with the known Aa of the mother, you will see that there are 4 of 16 possibilities resulting in aa. Thus, the probability of being aa are 1/4 ... and if 1/20 of those have the disease, then the result is 1/80.