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I am aware that The outcomes for two kids can be {BB, BG, GB, GG}, and given that there's one girl, the answer would be 1/3. But, when I put the question into a context of Bayesian Theorem, I seem to struggle.

My equation is as follows: P(2G|1G)=[P(1G|2G)*P(2G)]/P(1G)

where P(1G|2G) = 1 as given there are two girls, the probability of having 1 girl would be 1, P(2G)=0.5*0.5, P(1G)=0.5 .

Where, in my equation, did I get wrong?

nicole hi
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    Your question is ambiguously phrased. It matters why we know that Alice has a girl child. Regardless, this question has been discussed ad nauseum on this site already. – JMoravitz Aug 26 '19 at 11:00
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    The mistake for the traditional approach and interpretation of the problem that you made was in saying $P(1G)=0.5$. This should be $0.75$ instead. $P(1G)$ is not the probability that the "first" child is a girl., rather $P(1G)$ is the probability that at least one of the two kids that we know Alice has is a girl that we assume various things about to make the problem work such as independence of gender, etc... – JMoravitz Aug 26 '19 at 11:02
  • Go through the possible cases to see the flaw. There are three possibilities for the two kids leading to at least one girl : girl-girl , boy-girl , girl-boy. One of the three possibilities leads to two girls. – Peter Aug 26 '19 at 11:03
  • @JMoravitz Of course, we must assume that each combination is equally like , but in the exercise we can assume this. – Peter Aug 26 '19 at 11:04

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Your method is correct but you must remember that your P(1G) is the probability of at least one girl i.e. $3/4$.

Then your answer will be correct as well as your method!