section_path: Mathematical treatment/Not (necessarily) mutually exclusive events
section_name: Not (necessarily) mutually exclusive events
paragraph_text: Not (necessarily) mutually exclusive events
  
  If the events are not (necessarily) mutually exclusive then P (A or B) = P (A ∪ B) = P (A) + P (B) − P (A and B). {\displaystyle P\left(A{\hbox{ or }}B\right)=P(A\cup B)=P\left(A\right)+P\left(B\right)-P\left(A{\mbox{ and }}B\right).} Rewritten, P (A ∪ B) = P (A) + P (B) − P (A ∩ B) {\displaystyle P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)}
  
  For example, when drawing a card from a deck of cards, the chance of getting a heart or a face card (J, Q, K) (or both) is 13 52 + 12 52 − 3 52 = 11 26, {\displaystyle {\tfrac {13}{52}}+{\tfrac {12}{52}}-{\tfrac {3}{52}}={\tfrac {11}{26}},} since among the 52 cards of a deck, 13 are hearts, 12 are face cards, and 3 are both: here the possibilities included in the "3 that are both" are included in each of the "13 hearts" and the "12 face cards", but should only be counted once.
  
  This can be expanded further for multiple not (necessarily) mutually exclusive events. For three events, this proceeds as follows: P (A ∪ B ∪ C) = P ((A ∪ B) ∪ C) = P (A ∪ B) + P (C) − P ((A ∪ B) ∩ C) = P (A) + P (B) − P (A ∩ B) + P (C) − P ((A ∩ C) ∪ (B ∩ C)) = P (A) + P (B) + P (C) − P (A ∩ B) − (P (A ∩ C) + P (B ∩ C) − P ((A ∩ C) ∩ (B ∩ C))) P (A ∪ B ∪ C) = P (A) + P (B) + P (C) − P (A ∩ B) − P (A ∩ C) − P (B ∩ C) + P (A ∩ B ∩ C) {\displaystyle {\begin{aligned}P\left(A\cup B\cup C\right)=&P\left(\left(A\cup B\right)\cup C\right)\\=&P\left(A\cup B\right)+P\left(C\right)-P\left(\left(A\cup B\right)\cap C\right)\\=&P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)+P\left(C\right)-P\left(\left(A\cap C\right)\cup \left(B\cap C\right)\right)\\=&P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(A\cap B\right)-\left(P\left(A\cap C\right)+P\left(B\cap C\right)-P\left(\left(A\cap C\right)\cap \left(B\cap C\right)\right)\right)\\P\left(A\cup B\cup C\right)=&P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(A\cap B\right)-P\left(A\cap C\right)-P\left(B\cap C\right)+P\left(A\cap B\cap C\right)\end{aligned}}} It can be seen, then, that this pattern can be repeated for any number of events.