Given that at least one is a girl with characteristic C, show that the probability that both children are girls $= (2−p)/(4−p)$. This is Blitzstein, Introduction to Probability (2019 2 ed) Ch 2, Exercise 29, p 88.

A family has two children. Let C be a characteristic that a child can have, and assume
that each child has characteristic C with probability p, independently of each other and
of gender. For example, C could be the characteristic “born in winter”. Show that the probability that both children are girls given that at least one
is a girl with characteristic C is $(2−p)/(4−p)$
, which is 1/3 if $p = 1$ and approaches 1/2 from below as $p \to 0$ (agreeing with Example 2.2.7).

Can somebody correct my wrong attempt below?
Define $2G$ ="Both are Girls", $A$ = "at least one girl".
$P(C)=p$, since C is independent from gender and each other. Then $P(C|A)=P(C)=p$.
I need to find $P(2G|A,C)=P(2G$ and $A$ and $C)/P(A$ and $C)$. I tried to calculate $P(A$ and $C)=P(C$ and $A)=P(C|A)P(A)$.
$P(A)=1/3, P(C|A)=p \implies P(C$ and $A)=p/3=P(L)$.
I define $L = C \cap A$ to simplify  notation. Then
$P(2G|A,C)=P(2G|L)=P(2G$ and $L)/P(L)=P(L|2G)P(2G)/P(L)$.
In my new notation, $P(L|2G)$ is the probability that "at least one girl has C" given that both are girls. I tried to calculate  $P(L|2G)$  based on common sense, by dividing it into two cases where:

*

*$P($one $C|2G)$. Thus, only one of them has C $\implies$ $P($one $C|2G)$=first has & second not + first not & second has $= pq+qp=\color{limegreen}{2pq}$, where $q=1-p$.


*$P($both have $C|2G)$. Since C is independent from gender and each other, then $P($both have $C|2G)=\color{red}{pp}$.
Hence, $P(L|2G)=\color{limegreen}{2pq}+\color{red}{pp}=2p(1-p)$.
I think $P(2G) = 1/4$, since there are only 4 choices:  BB,  GG, GB, BG.
I concluded $P(2G|L)=\dfrac{2p(1-p)\times 1/4}{p/3}=\dfrac{3(1-p)}{2}$. But this is certainly wrong.
 A: A child is $~\boxed{\begin{array}{rl}\text{girl with C} & \tfrac 12 p\\ \text{girl without C} & \tfrac 12(1-p)\\ \text{a boy} & \tfrac 12\end{array}}$ 
Two children are:  $~\boxed{\begin{array}{rl}\text{2 girls with C} & \tfrac 14 p^2 & \star\\ \text{a girl with and a girl without, C}& \tfrac 24p(1-p) &\star \\ \text{2 girls without C} & \tfrac 14(1-p)^2\\ \text{a girl with C, a boy} & \tfrac 24 p & \ast \\ \text{a girl without C, a boy} & \tfrac24 (1-p)\\ \text{2 boys}& \tfrac 14\end{array}}$
Reality Check: The probabilities in the table sum to $1$.
The probability that both children are girls and at least one has C: ...
The probability that at least one child is girl with C: ...
The probabiltiy that both children are girls given at least one is a girl with C: ...
A: I will finish Graham Kemp's answer.
The final answer depends on the 3 probabilities colored below.
Two children are:    $~\boxed{\begin{array}{rl}\color{limegreen}{\text{2 girls with C}} & \color{limegreen}{\tfrac 14 p^2} & \star\\ \color{red}{\text{a girl with and a girl without, C}} & \color{red}{\tfrac 24p(1-p)} &\star \\ \text{2 girls without C} & \tfrac 14(1-p)^2\\ \color{lightskyblue}{\text{a girl with C, a boy}} & \color{lightskyblue}{\tfrac 24 p} & \ast \\ \text{a girl without C, a boy} & \tfrac24 (1-p)\\ \text{2 boys}& \tfrac 14\end{array}}$.
Then by Bayes's Theorem, P(both children are girls | $\ge 1$ is a girl with characteristic C) = P(both children are girls & $\ge 1$ is a girl with characteristic C)/P($\ge 1$ is a girl with characteristic C) = $\dfrac{\color{limegreen}{\tfrac 14 p^2} + \color{red}{\tfrac 24p(1-p)}}{\color{limegreen}{\tfrac 14 p^2} + \color{red}{\tfrac 24p(1-p)} + \color{lightskyblue}{\tfrac 24 p}} = \dfrac{2p - p^2}{4p - p^2}$. QED.
