$2$-girl problem with bias of $P(G) = 0.75$ and $P(B) = 0.25$ A family has 2 children. The probability of a child being a girl is 0.75. We pick one of them at random and find out that she is a girl. What is the probability that all their children are girls?

My solution : It is given that $\text{P(child is a girl)}$ =  $\text{P(G)} = 0.75$
$\text{P(B)} = 0.25$
There are four possibilities in this family : $\{ \text{BB, GG, BG, GB} \}$
P(BB) = $0.25 \times 0.25$ 
P(GG) = $0.75 \times 0.75$ 
P(GB) = $0.75 \times 0.25$ 
P(BG) = $0.25 \times 0.75$
Now question mentions that we pick one of them at random and find out that she is a girl. Hence we can reduce the sample space by removing the case P(BB).
Sample space is now $\{ \text{ GG, BG, GB} \}$
$\text{P(GG | one of the child is a girl)}$ = $\Large \frac{0.75^2}{0.75^2 + 2 \times (0.75 \times 0.25) }$
$\text{P(GG | one of the child is a girl)} = 0.60$

The given answer in my university quiz is 0.67.
Explanation: 
The sample space is Ω = {BB, BG, GB, GG}
Let G_r be the event that a randomly chosen child is a girl 
From the data given, 
$P(G) = 0.75$ and $P(B) = 0.25$
So, $P(GG)=9/16, P(GB)=P(BG)=3/16, P(BB) = 1/16$ 
Now, 
$P(G_r|BB) = 0$ 
$P(G_r|BG) = P(G_r|GB) = 0.75$ 
$P(G_r|GG) = 1$ 
We would like to find $P(GG|G_r)$ 
$P(GG|G_r) = $$\large \frac{P(G_r|GG)P(GG)}{P(G_r)}$ 
$\Large = \frac{1.\frac{9}{16}}{P(G_r|BB)P(BB) + P(G_r|BG)P(BG) + P(G_r|GB)P(GB) + P(G_r|GG)P(GG)}$
$\Large  = \frac{\frac{9}{16}}{0.\frac{1}{16} + 0.75.\frac{3}{16} + 0.75.\frac{3}{16} + 1.\frac{9}{16}}$
Solving, we will get $P(GG|G_r) = \frac{2}{3} = 0.67$
 A: There are two kids and we randomly pick one.
It appears to be a girl.
This tells us nothing at all about the gender of the other kid.
The only relevant info concerning the question "are both girls?" we get is that this will indeed be the case if and only if the other kid is a girl.
The probability of that event is 0.75.

Edit:
Your mistake: the condition is not the same as the "at least one of the kids is a girl".
Mistake of quiz: $P(G_r|GB)$ and $P(G_r|BG)$ do not equalize $0.75$.
If you pick randomly from a boy and the girl then the probability of picking a girl is evidently $0.5$.
If this correct value is substituted then it gives $0.75$ as final answer (as it should).
IMV invoking the rule of Bayes here is not necessary, is cumbersome and is complicating.
A: Suppose a family has two children and we pick a child at random.  Our sample space is:

*

*GG, pick first child

*GG, pick second child

*GB, pick first child

*GB, pick second child

*BG, pick first child

*BG, pick second child

*BB, pick first child

*BB, pick second child

If we assume that we had an equal chance of picking the older and younger children, and with the given probabilities for each child to be a gender, we have the probabilities for each of the eight of:

*

*$\frac{3}{4}\frac{3}{4}\frac{1}{2}=\frac{9}{32}$

*$\frac{3}{4}\frac{3}{4}\frac{1}{2}=\frac{9}{32}$

*$\frac{3}{4}\frac{1}{4}\frac{1}{2}=\frac{3}{32}$

*$\frac{3}{4}\frac{1}{4}\frac{1}{2}=\frac{3}{32}$

*$\frac{1}{4}\frac{3}{4}\frac{1}{2}=\frac{3}{32}$

*$\frac{1}{4}\frac{3}{4}\frac{1}{2}=\frac{3}{32}$

*$\frac{1}{4}\frac{1}{4}\frac{1}{2}=\frac{1}{32}$

*$\frac{1}{4}\frac{1}{4}\frac{1}{2}=\frac{1}{32}$
Now, we are told that when we picked a random child, we picked a girl.  This rules out cases 4,5,7, and 8
We want:
$$\begin{align}P(GG|chose\,girl)&=\frac{P(1)+P(2)}{P(1)+P(2)+P(3)+P(6)}\\&=\frac{\frac{9}{32}+\frac{9}{32}}{\frac{9}{32}+\frac{9}{32}+\frac{3}{32}+\frac{3}{32}}\\&=\frac{9+9}{9+9+3+3}\\&=\frac{18}{24}\\&=\frac{3}{4}\end{align}$$
