# Proof

(related to Proposition: Probability of Event Union)

We first note that

$A\cup B=(A\cap B)\cup(A\cap\overline B)\cup (\overline A\cap B).$

Because the events $$(A\cap B)$$, $$(A\cap\overline B)$$ and $$(\overline A\cap B)$$ are mutually exclusive, it follows from the definition of probability that

$\begin{array}{ccl} p(A\cup B)&=&p(A\cap B)+p(A\cap\overline B)+p(\overline A\cap B)\\ &=&p(A\cap(B\cup \overline B))+p(\overline A\cap B)\\ &=&p(A\cap\Omega)+p(\overline A\cap B)\\ &=&p(A)+p(\overline A\cap B)\\ \end{array}~~~~~~~~~~~( * ).$

Because $$B=(A\cap B)\cup (\overline A\cap B)$$, and because again, both events $$(A\cap B)$$ and $$(\overline A\cap B)$$ are mutually exclusive, we have $p(\overline A\cap B)=p(B)-p(A\cap B).$

Together with $$( * )$$, we get

$p(A\cup B)=p(A)+p(B)-p(A\cap B),$

as required.

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### References

#### Bibliography

1. Bosch, Karl: "Elementare Einführung in die Wahrscheinlichkeitsrechnung", vieweg Studium, 1995, 6th Edition