Q:

Probability (Bayes Rule)

One quarter of twins are identical. Identical twins are always of the same sex. With fraternal (= non-identical) twins, the sexes are independent 50-50, like tossing a fair coin twice.
What fraction of same-sex twins are identical?
I know that this is a conditional probability question and I understand how conditional probability works. But, how can I get the answer when I do not know the percentage of same-sex and different-sex twins for identical twins?

A:

$Pr(A\mid B)=\dfrac{Pr(A\cap B)}{Pr(B)}=\dfrac{Pr(B\mid A)Pr(A)}{Pr(B)}$
Let $A$ represent that the twins are identical.  Let $B$ represent that twins are samesex.  The problem tells us $Pr(A)=\frac{1}{4}$, that $Pr(B\mid A)=1$ and we can show that $Pr(B\mid A^c)=\frac{1}{2}$ due to the independence and uniformity assumption.
We have then:
$$Pr(B)=Pr(A\cap B)+Pr(A^c\cap B) = Pr(A)Pr(B\mid A)+Pr(A^c)Pr(B\mid A^c)$$
$$=\frac{1}{4}\cdot 1+\frac{3}{4}\cdot\frac{1}{2}=\frac{5}{8}$$
We are asked to find $Pr(A\mid B)$
$$Pr(A\mid B)=\frac{Pr(B\mid A)Pr(A)}{Pr(B)}=\frac{1\cdot \frac{1}{4}}{\frac{5}{8}}=\frac{2}{5}$$

