Question: Duplicate the method of proof used in Exercise 2.12 to show that P(A B C D) = P(A) + P(B) + P(C)
P(A ∪ B ∪ C ∪ D) = P(A) + P(B) + P(C) + P(D)
- P(A ∩ B) - P(A ∩ C) - P(A ∩ D)
- P(B ∩ C) - P(B ∩ D) - P(C ∩ D)
+ P(A ∩ B ∩ C) + P(A ∩ B ∩ D)
+ P(A ∩ C ∩ D) + P(B ∩ C ∩ D)
- P(A ∩ B ∩ C ∩ D)
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