$\textbf{The Problem:}$ 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.

  • 1
    $\begingroup$ 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. $\endgroup$ – lulu Feb 14 at 2:15
  • $\begingroup$ @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 :) $\endgroup$ – G the Stackman Feb 14 at 2:18
  • 1
    $\begingroup$ 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. $\endgroup$ – lulu Feb 14 at 2:23
  • 2
    $\begingroup$ 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 $$P(GG) = P(GG\mid I)P(I) + P(GG\mid F)P(F) = \frac{1}{2}\frac{1}{3}+\frac{1}{4}\frac{2}{3}=\frac{1}{3}.$$ $\endgroup$ – spaceisdarkgreen Feb 14 at 2:29
  • 1
    $\begingroup$ @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. $\endgroup$ – G the Stackman Feb 14 at 2:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.