$\textbf{The Problem:}$ Assume that $\frac{1}{3}$ of all twins are identical twins. You learn that Miranda is expecting twins, but you have no other information. Explain any assumptions you may make below.
$\textbf{a)}$ Find the probability that Miranda will have two girls.
We can set up a probability space for this experiment, which agrees with the assumption, as follows. Define the sample space to be $$\Omega=\{(gg),(bb),(bg),(gb),(GG),(BB)\},$$ where $g$ stands for girl and $b$ for boy in the fraternal twins case, and $G$ and $B$ stands for girl and boy, respectively, in the identical twins case. Take $\mathcal F=2^\Omega$ and let $P(A)=\frac{|A|}{|\Omega|}$ for all $A\in\mathcal F$, where we assume all outcomes to be equalliy likely. Now let $I$ be the event of identical twins and $F$ that of fraternal twins. Now, in accordance with the assumption, we have $P(I)=\frac{1}{3}$ and $P(F)=\frac{2}{3}$. Let $G$ be the event of having two girls. It follows that $$P(G)=\frac{|G|}{|\Omega|}=\frac{2}{6}=\frac{1}{3}.$$
$\textbf{b)}$ You learn that Miranda gave birth to two girls. What is the probability that the girls are identical twins?
Under the same assumptions as in part (a), we have that $$P(I|\text{two girls})=\frac{P(I\cap\text{two girls})}{P(\text{two girls})}=\frac{1/6}{1/3}=\frac{1}{2}.$$
$\textbf{My Concerns:}$ Do you agree with my approach above? My main concern is whether setting up a probability space which agrees with the assumptions and then carrying out the calculations is allowed. I have seen another solution using conditional probabilities which in the end seems to have the same underlying space and measure, and assumptions.
Thank you for your time and I appreciate any feedback.