Problem source

Every person  carries two genes which can each be either of 
 type $A$ or of type $B$. 
It is known that $81\%$ of the population are $AA$ (i.e. both genes are 
of type $A$), $18\%$ are $AB$ (i.e. there is one gene of type $A$ 
and one of type $B$) and $1\%$ are $BB$. A child inherits 
one gene from each of its parents. If one parent is $AA$, the child
 inherits a gene of type  $A$ from that parent;  
if the  parent is $BB$, the child
 inherits a gene of type  $B$ from that parent; 
 if the parent 
is $AB$, the inherited gene is equally likely to be $A$ or $B$. 
\begin{questionparts}
\item
Given that two $AB$ parents have four children, 
show that the probability 
that two of them are $AA$ and two of them are $BB$ is $3/128$. 
 
\item
My mother is $AB$ and I am $AA$. 
Find the probability that my father is $AB$.
\end{questionparts}