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Here's a question I'm thinking about:

There are three drawers. Drawer A contains 2 black socks, Drawer 2 contains two white socks and the third drawer is mixed. If I pulled a sock from one of the drawers. Given that it is white, what is the probability that the other sock in the drawer is white?

So what I'm thinking is that I picked either the second or the third. If I picked the second I know the other one is white, and if I picked the third I know that the other one is black and so the probability is $1\cdot \frac{1}{2} + 0\cdot \frac{1}{2}=\frac{1}{2}$

I saw someone else's solution and he says that after picking one sock we have a new sample space $$\Omega=\{(W_1,W_2),(W_2,W_1),(W_m,B)$$ where $W_1,W_2$ are the two in the second drawer and $W_m$ is the one in the mixed drawer. Anyway he concluded that the probability is $$P(second\, is\, white)=\frac{\{(W_1,W_2),(W_2,W_1)\}}{|\Omega|}=\frac{2}{3}$$

I think he's mistaken for counting the order in which the socks were pulled, because it is given that we pulled a white one, it doesn't matter which exactly.

So, who is right? And if I'm right, was I also right about his mistake?

Thanks!

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    You mean "Drawer A contains $2$ black socks"?2012-11-12
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    "What is the probability that the other sock in the drawer is white?" is a strange question, since the sock's colour is deterministic.2012-11-12
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    @Cocopuffs: That's only under some interpretations of probability. I would argue that an interpretation of probability that doesn't allow us to speak of a probability in this case offers an unduly restricted concept of probabilities.2012-11-12
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    @joriki: yes that's what I meant2012-11-12
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    @joriki Yes, that's true. In any case, it removes the 'paradox': if you ask yourself what the probability a randomly chosen drawer will not be mixed is, this probability can't change by picking socks later on. As Hagen says.2012-11-12
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    By the way, this is a well-known problem: http://en.wikipedia.org/wiki/Bertrand's_box_paradox2012-11-12

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