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If I want to find $P(A \cap B)$, is it $1-P(A^\complement \cap B^\complement)$ or $1-P(A \cap B)^\complement$?

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A compliment is a friendly remark. The set of elements not contained in a set is its complement. – joriki Oct 13 '12 at 20:43
So it is! Fixed. Thanks. – karoma Oct 13 '12 at 20:44
$P(A \cap B)^\complement$ doesn't make sense, $P(A \cap B)$ is a real number from $[0,1]$. – fgp Oct 13 '12 at 20:47
A Venn diagram will let you quickly identify $A^c\cap B^c$ and $(A\cap B)^c$ visually. Then you can see that $1-P((A\cap B)^c)$ is right, and also see why the other is in general not right. – André Nicolas Oct 13 '12 at 21:11

The complement of $A\cap B$ is $(A\cap B)^c$, so $P(A\cap B)=1-P\big((A\cap B)^c\big)$. Using one of the de Morgan’s laws you can go further: $(A\cap B)^c=A^c\cup B^c$.

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