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I got a little stuck on a simple proof of the following probability identity.

Given

$\mathbb{P}(A^c \cap B^c)=1-\mathbb{P}(A)-\mathbb{P}(B)+\mathbb{P}(A\cap B)$

how to prove for any set $X$,

$\mathbb{P}(X \cap A^c \cap B^c)=\mathbb{P}(X)-\mathbb{P}(X\cap A)-\mathbb{P}(X\cap B)+\mathbb{P}(X\cap A\cap B)$

Looks very intuitive; just replace the whole space by $X$. But how to prove it simply and rigorously? Thanks.

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    Why would you want to deduce the latter from the former? The latter can be proven the same way as the former.2011-02-12
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    @Rasmus: oh, I want to know the details about the proof. I thought derivation from the former is the only way to get the latter. Thanks.2011-02-12
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    If you have seen conditional probabilities, your intuition can be easily formalized.2011-02-12
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    @Raskolnikov: how to argue with conditional probability?2011-02-12
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    Just think about it and remember that $\mathbb{P}(A \cap X) = \mathbb{P}(A | X) \mathbb{P}(X)$.2011-02-12
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    @Raskolnikov: It is still very easy to do this from conditional probability point of view, unless you can give the complete formalized argument. And if so, I would accept your answer. :D2011-02-13

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