Variation on the classic Boy - Girl Probability Problem Revisiting the pair Boy-Girl probability question that I got asked during an interview ~6 months ago.

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*The first question is similar to the classic "if the family has 2 children and 1 girl, what is the probability the 2nd child is a also a girl", however, I specifically asked the interviewer if we could assume that we know that a daughter came first


*The follow up question is what stumped me - repeat the question above, however, assume the probability of having a girl is now 2/3.
For the first question, I answered 1/2. However, I wasn't expecting a follow-up so I believe I just overthought the 2nd question in the heat of the interview.. but the answer I came up with was 2/3. Wondering if I got the right answer here (evidently messed up somewhere, I didn't get the job)
 A: The question is kind of ambiguous, I think it appears as a problem in David Mackay's classic inference text. Anyway, let me rephrase your question as "If the number of girls among the children is greater than 1, what is the probability that both children are girls." I think this is what the question intends.
This is a straightforward application of conditional probability if we clearly define our events. We let the marginal probability of having a girl be denoted by $p$.
$$ P(\# G = 2 \vert \# G \geq 1) = \frac{P(\#G = 2, \#G \geq 1)}{P(\# G \geq 1)}$$
The two events $(\#G = 2, \#G \geq 1)$ are already contained in the event $\#G = 2$, so the numerator is just $p(\# G = 2) = p^2$.
The event $\#G \geq 1$ is the complement of the event of having no female children, so $p(\#G \geq 1) = 1 - p(\#G = 0) = 1 - (1-p)^2 $. So this simplifies to:
$$ P(\# G = 2 \vert \# G \geq 1) = \frac{p}{2 - p}$$.
So for the first case assuming equal prior probabilities $p=1/2$ and the answer is $1/3$, but for the second case you're given $p=2/3$ and $ P(\# G = 2 \vert \# G \geq 1) = 1/2$.
