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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 $P(A\cap B)=P(B|A)\cdot P(A)=\frac{1}{3}\cdot \frac{1}{4}=\frac{1}{12}.$


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.

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    $\begingroup$ It lacks information on twins. I have made an assumption on the probability of births and hopefully it is right. $\endgroup$ – Satish Ramanathan Feb 15 at 11:47
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    $\begingroup$ You assumed that $\frac{1}{3}$ of twins which are girls are identical twins, but the problem statement said that $\frac{1}{3}$ of all twins are identical twins. Do you see how that's not the same thing? $\endgroup$ – David K Feb 15 at 12:44
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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?

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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 $\frac{p^2}{3}$.

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  • $\begingroup$ sorry, i dont understand this proof. what is your $p = $ "probability that a boy or girl is born"? if $p= $ Prob(a boy-or-girl is born), then $p=1$. if instead you mean $p=$Prob(a boy is born) = Prob(a girl is born), then $p = 1/2$. Note: the question states as a given that Miranda is going to have twins, i.e. she is going to give birth to 2 babies, not just 1. $\endgroup$ – antkam Feb 16 at 17:54
  • $\begingroup$ You are right. It is prob( a boy is born) = prob( a girl is born) = p. Further assumption is a twin is born ( boy-boy, boy-girl, girl-girl) all equal to $p^2$ and thus the equation $1 = 2p+3p^2$. Now it must be clear. Thanks $\endgroup$ – Satish Ramanathan Feb 17 at 10:47
  • $\begingroup$ it makes no sense to assume $p^2 =$ Prob(a twin is born). also, the question already told you Miranda is giving birth to twins. $\endgroup$ – antkam Feb 17 at 15:26

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