Conditional Probability Problem about identical twins

Question

Assume that $\frac{1}{3}$ of all twins are identical twins. You learn that Miranda is expecting twins, but you have no other information.

a) Find the probability that Miranda will have two girls.

b) You learn that Miranda gave birth to two girls. What is the probability that the girls are identical twins?

For a), we have, letting $A$ be the event that Miranda will have two girls, and assuming that the probability a child is a girl at birth is $\frac{1}{2},$ we have $P(A)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}.$

For b), letting $B$ be the event that the girls are identical twins, then we find that $P(A\cap B)=P(B|A)\cdot P(A)=\frac{1}{3}\cdot \frac{1}{4}=\frac{1}{12}.$

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Is the above reasoning correct? I think that I have assumed that $\frac{1}{3}$ of twins which are girls are identical twins. Would that be incorrect, since it is also stated to explain any assumptions one makes?

Thank you for your time and appreciate the feedback.

Answer 1

HINTS only

For ease of notation name these events:

• $G=$ having 2 girls

• $I=$ having identical twins

• $F=$ having non-identical (i.e. fraternal) twins; note that $F$ is the complement of $I$.

(a) is asking for $P(G) = P(G | I) P(I) + P(G | F) P(F)$. Here, $P(I) = {1 \over 3}$ is given, and therefore $P(F) = 1 - P(I) = {2 \over 3}$. Can you make reasonable assumptions about the other two terms? Can you finish from here?

(b) is asking for $P(I | G) = { P(I \cap G) \over P(G)} = {P(G|I)P(I) \over P(G)}$. Can you finish from here?

Answer 2

Let $p$ be the probability that a boy or girl is born. Assume that you have $p^2$ as the probability that a twin is born.

Then $1 = 2p+3p^2 \implies p = \frac{1}{3}$

The probability that she will have a girl twin is $p^2$.

The probability that she will have an identical twin $\frac{p^2}{3}$.