# Probability Problem with Twins

$$\textbf{The Problem:}$$ 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.

• Are we to assume that opposite sex fraternal twins and same sex fraternal twins are equally probable? I believe that this is quite far from true so if you want it as an assumption to make a math problem out of it, you should make the assumption explicit. Similarly for assumptions of gender distribution in the two cases. – lulu Feb 14 at 2:15
• @lulu Would that not be included when I mention that I assume all outcomes in my sample space to be equally likely? And thank you very much for your feedback :) – G the Stackman Feb 14 at 2:18
• Your sample space is meant to encode the assumptions of the problem, but those assumptions should be made explicitly, especially when they aren't accurate. I forget the exact numbers, but I think I've seen things like $60\%$ for opposite sex fraternal twins. – lulu Feb 14 at 2:23
• It's unnecessarily confusing to say they're all equally likely... it makes it seem like you're making bad assumptions since they wouldn't be equally likely if the probability of identical twins weren't a magic number $1/3$. Instead, just say $$P(GG) = P(GG\mid I)P(I) + P(GG\mid F)P(F) = \frac{1}{2}\frac{1}{3}+\frac{1}{4}\frac{2}{3}=\frac{1}{3}.$$ – spaceisdarkgreen Feb 14 at 2:29
• @spaceisdarkgreen I see what you mean. I changed the assumption to the probability of identical twins been $1/4$, and as you said, it breaks down. Thanks for your help. – G the Stackman Feb 14 at 2:34