For events A and B it is known that P(A) = 1/2, P(B) = 1/3 and P(A ∪ B) = 3/4. Find P(A ∩ B).

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Missing intersection probability

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For events $A$ and $B$ it is known that $P (A) = \frac{1}{2}$ , $P (B) = \frac{1}{3}$ and $P (A \cup B) = \frac{3}{4}$ . Find $P (A \cap B)$ .

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Hint

Use the inclusion-exclusion formula for a union of $2$ events.

Solution

Inclusion-exclusion gives $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ .
Hence $P (A \cap B) = \frac{1}{2} + \frac{1}{3} - \frac{3}{4}$ .
With common denominator $12$ this is $\frac{6}{12} + \frac{4}{12} - \frac{9}{12}$ .
The result is $\frac{1}{12}$ .

Answer: $\frac{1}{12}$