On proving $P(A^c \cap B) = P(B) - P(A \cap B)$

Question

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

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.
I'd have posted an answer, but "cardinal"'s comment says essentially what I would have said.
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?
@MichaelHardy: You (or others) should feel free to post an answer. Cheers. :)
@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.

Listing · Answer 1 · 2011-12-11 00:23:47Z

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.

1 Answer