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My initial thought process:
Sample space: GG, GB, BG, BB. I then crossed out BG because it's the same as GB because order doesn't matter here. And because we know one is a girl, there leaves two possibilities left. If the other is a boy, the probability should be 1/2, much like how they deduced it Finding probability of other child also being a boy.

However, I was told by my teacher that the answer is not 1/2. I'm wondering if any of you guys can see a way in how the question is worded so that it's not 1/2... I don't think there's any other factors?

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1 Answer

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You should not have omitted the BG!

The sample space is (and needs to be) $\{BB, BG, GB, GG\}.$

We know one child is a girl, so that rules out BB. That leaves us with a sample space of BG, GB, GG

In which case the probability that the second child is a boy is $\dfrac 23$.

Each of the four outcomes has a probability of $\frac 14$. BG: "Having a boy, and then a girl" is a different outcome than $GB:$ having a girl, and then a boy.

To omit one of the boy-girl/girl-boy pairs leaves a sample space of three, with each outcome having probability of 13, which is not correct. Having a boy-girl pair is twice as likely as having two boys, and twice as likely as having 2 girls, and we can only obtain this by counting all four outcomes as distinct, indeed, distinguishable.

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