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In my introductory statistics class one of the problems was to determine $P(A \cap B)$ given $P(A \cup B)$, $P(A)$ and $P(B)$.

Using $$P(A \cap B) = P(A) + P(B) - P(A \cup B)$$ we can solve:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Now, as a matter of curiosity, I'm wondering how to express this idea in terms of the sets themselves and not in terms of their probabilities.

In other words, for two sets $A$, $B$, and $C = A \cup B$, how do we express $A \cup B$ in terms of their union $C$? The union operator doesn't mind duplicates, but the equation does.

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2 Answers 2

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If we define the cardinality of a set $|A|$ as its number of elements then for finite sets: $|A \cup B| = |A|+|B|-|A \cap B|$

For some quick intuition, if we consider $A, B$ as subsets of a sample space then their normalized cardinality with respect to the sample space can be used as a probability function (with a loss of generality). Then $P(A) = \frac{|A|}{|\Omega|}$ where $\Omega$ is the sample space.

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In general, if $\mathbb P$ is a probability measure on a measurable space $(\Omega, \mathcal F$) and $A,B\in\mathcal F$, then by countable additivity we have

\begin{align} \mathbb P(A)+\mathbb P(B) &= \mathbb P((A\cap B)\cup (A\cap B^c)) + \mathbb P((B\cap A)\cup (B\cap A^c))\\ &= \mathbb P(A\cap B) + \mathbb P(A\cap B^c) + \mathbb P((B\cap A)\cup (B\cap A^c))\\ &= \mathbb P(A\cap B) + \mathbb P((A\cap B^c)\cup(B\cap A)\cup (B\cap A^c))\\ &= \mathbb P(A\cap B) + \mathbb P(A\cup B), \end{align} from which we conclude.