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Lesson 27 · Either, or, or both
The overlap gets counted twice
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The overlap gets counted twice

Two events, and the word or. Write A∪BA \cup B for A or B, or both — the cup-shaped symbol is the union, and it is generous: it accepts anything that satisfies either event. Its partner A∩BA \cap B is the intersection, read A and B, the outcomes where both hold at once.

If the two events cannot possibly happen together — mutually exclusive — the shares simply add: P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B). If they CAN overlap, that sum charges the overlap twice, once inside A and once inside B, so it gets refunded exactly once: P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B), read aloud P of A-union-B equals P of A plus P of B minus P of A-intersect-B. Draw two overlapping circles; the lens in the middle is the whole lesson.

The exclusive rule is not a different law, it is this one with a zero in it. And the relation has four slots, so it runs backwards too: hand it any three and it returns the fourth — which is how a question about an unknown OVERLAP is answered from a known union. Keep the range check armed throughout. A union that comes out above 1 means the overlap never came off.

P(A)P(B)P(A∪B)

P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B)

  • P(A∪B)P(A \cup B)= Probability of A or B
  • P(A)P(A)= Probability of A
  • P(B)P(B)= Probability of B
Addition Rule (Mutually Exclusive Events) solver →
P(A)P(B)P(A∩B)

P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

  • P(A∪B)P(A \cup B)= Probability of A or B
  • P(A)P(A)= Probability of A
  • P(B)P(B)= Probability of B
  • P(A∩B)P(A \cap B)= Probability of A and B
General Addition Rule solver →