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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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iamashortgiraffe
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1 Answers1

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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.

amWhy
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  • I was thinking about that, but why? You're just taking a pair of children, so whether it's BG or GB shouldn't matter because either way you're getting a girl and boy? – iamashortgiraffe Dec 14 '13 at 22:08
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    Because each of the four outcomes has a probability of $\dfrac 14$. Having a boy, and then a girl is a different outcome than having had a girl and then a boy. To omit one of the boy-girl pairs leaves a sample space of three, with each outcome having probability of $\dfrac 13$, 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. – amWhy Dec 14 '13 at 22:13
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    @Adee: It matters because the two possibilities are distinguishable. If you knew that the older child was a girl, then the sample space would be simply ${GG,GB}$, where the older child is listed first, and the probability would indeed be $\frac12$. Similarly, if you knew that the younger child was a girl, the probability would be $\frac12$. If you knew that the taller child was a girl, it would be $\frac12$. But all you know here is that one of the children is a girl; if you list the possibilities in the order older-younger, you can rule out only $BB$. – Brian M. Scott Dec 14 '13 at 22:15
  • Thanks for chiming in, @Brian. The word "distinguishable" was on the tip of my tongue, but I couldn't "find" it when I was trying to explain in the comments. I was toying with the older-younger frame, as well, or the analogy between tossing two coins in a row: first coin, second coin... – amWhy Dec 14 '13 at 22:17
  • @amWhy: Needs another TU +1 – Amzoti Dec 15 '13 at 00:25