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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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Hint: What is $(A^c \cap B) \cap (A \cap B)$? What is $A^c \cup A$? Moving the $P(A \cap B)$ term to the other side may make things seem clearer. –  cardinal Dec 10 '11 at 16:53
    
I'd have posted an answer, but "cardinal"'s comment says essentially what I would have said. –  Michael Hardy Dec 10 '11 at 16:57
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A different viewpoint (but basically the same thing as the above comments): Do you know that $B=(A^c\cap B) \cup (A\cap B)$ and the set $(A^c\cap B)$ and $(A\cap B)$ are disjoint? What do you know about probability of union of disjoint events? –  Martin Sleziak Dec 10 '11 at 16:58
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@MichaelHardy: You (or others) should feel free to post an answer. Cheers. :) –  cardinal Dec 10 '11 at 17:03
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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 that I'd encourage you to read over. Cheers. –  cardinal Dec 10 '11 at 17:18
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1 Answer 1

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.

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