Question on conditional probability with extra condition In a population $40$% of the children have blue eyes and $60$% have green eyes.
Secondly (the extra condition), if a child that is randomly selected has blue eyes then the probability for each other brother or sister of this child to have blue eyes as well is $75$%.

a) In this population a family having $2$ children is selected. What is the probability that both children have green eyes?
b) In this population a family having $2$ children of which at least one has green eyes is selected. What is the probability that both children have green eyes?

My try: for simplicity let it be that we are dealing with e.g. $100$  families that have $2$ children. Let $x$ of them have $BB$ (i.e. two children with blue eyes) let $y$ of them have $GB$ and also $y$ have $BG$ (same probability as $GB$) and let $z$ have $GG$. Then there are $2x+4y+2z=200$ children of which $2x+2y$ have blue eyes and $2y+2z$ have green eyes. This tells us that $2x+2y=0.4\times200=80$ and $2y+2z=0.6\times200=120$.
But I am stuck here and would not be surprised if this is not the route I should take. Can you help me with this?
Thanks in advance.

I have the following two-way table (probabilities):
$$\begin{array}{|m{cm}|m{1cm}|}
\hline  &  \text{child blue eyes}  &\text{child green eyes}   \\ \hline \hline \hline\text{broth./sist. blue eyes}& 0.3 (x)&0.1 &0.4\\ \hline \text{broth./sist. green eyes}& 0.1 (y)&0.5&0.6 \\ \hline & 0.4 &0.6 &1 \\ \hline  \end{array}$$ 
If a child that is randomly selected has blue eyes then the probability for each other brother or sister of this child to have blue eyes as well is 75%.
Thus $0.75=\frac{x}{0.4}\Longrightarrow x=0.75\cdot 0.4=0.3$
And $y=0.4-0.3=0.1$
A sister/brother is a child, too. So they have the same (unconditonal) probability to have blue eyes/green eyes. Now it is easy to complete the table.
Ad a) The answer can be read off the table.
Ad b)