<s>[INST] $2$-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 $P(GG|G_r) = \frac{2}{3} = 0.67$
 [/INST] 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:
$$\begin{align}P(GG|chose\,girl)&=\frac{P(1)+P(2)}{P(1)+P(2)+P(3)+P(6)}\\&=\frac{\frac{9}{32}+\frac{9}{32}}{\frac{9}{32}+\frac{9}{32}+\frac{3}{32}+\frac{3}{32}}\\&=\frac{9+9}{9+9+3+3}\\&=\frac{18}{24}\\&=\frac{3}{4}\end{align}$$
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