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This is embarrassing, but I am unable to prove that $P(A^c \cap B) = P(B) - P(A \cap B)$. Any pointers?

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    @PKR: For future reference in asking homework questions, many users will generally want to see some attempt at a solution or more developed reasoning explaining why and where you are stuck. There is a helpful [homework FAQ](http://meta.math.stackexchange.com/questions/1803/how-to-ask-a-homework-question) that I'd encourage you to read over. Cheers.2011-12-10

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What you want to show is equivalent to:

$P(A^c \cap B) + P(A \cap B) = P(B)$

Note by definition $(A^c \cap A)=\emptyset \Rightarrow (A^c \cap B) \cap (A \cap B) = \emptyset$

Besides you know that $(A^c \cap B) \cup (A \cap B)=B$

Therefore the statement follows by using the $\sigma$-additivity of $P$, namely

$(X \cap Y)=\emptyset \Rightarrow P(X \cup Y)=P(X)+P(Y)$

I let you write out the last step.