This is embarrassing, but I am unable to prove that $P(A^c \cap B) = P(B) - P(A \cap B)$. Any pointers?
On proving $P(A^c \cap B) = P(B) - P(A \cap B)$
3
$\begingroup$
probability
-
5**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. – 2011-12-10
-
0I'd have posted an answer, but "cardinal"'s comment says essentially what I would have said. – 2011-12-10
-
1A 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? – 2011-12-10
-
2@MichaelHardy: You (or others) should feel free to post an answer. Cheers. :) – 2011-12-10
-
2@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
1 Answers
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.