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For events A and B, probabilities P(A), P(B), P(A ∪ B) are given.

How do we calculate P(A ∩ B)?

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up vote 3 down vote accepted

Hint: Use the fact that $$\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B).$$ If you know three of these quantities, you can find the fourth.

Remark: The "fact" can be informally justified by drawing a Venn Diagram. When you add together the "weights" of $A$ and $B$, you are counting the weight of $A\cap B$ twice. So to find the weight of $A\cup B$, you should subtract the weight of $A\cap B$ from the sum of the weights of $A$ and of $B$.

For a more formal justification, note that $A\setminus B$, $A\cap B$, and $B\setminus A$ are pairwise disjoint, and their union is $A\cup B$. So $$\Pr(A\cup B)=\Pr(A\setminus B)+\Pr(A\cap B)+\Pr(B\setminus A).$$ It follows that $$\Pr(A\cup B)=\left(\Pr(A\setminus B)+\Pr(A\cap B)\right)+\left(\Pr(B\setminus A)+\Pr(A\cap B)\right)-\Pr(A\cap B).$$ The sum of the first two terms on the right is $\Pr(A)$, and the sum of the next two terms is $\Pr(B)$.

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