Question:
Prove that P(A union B union C ) = P(A) P(B) P(C) P(A intersection B) P(B intersection C ) P(A intersection B intersection C)
Answer:
Let D = B ∪ C
Now, P(A ∪ B ∪ C) = P(A ∪ D) = P(A) + P(D) - P(A ∩ D) ..............1
Now, A ∩ D = A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
So, P(A ∩ D) = P[(A ∩ B) ∪ (A ∩ C)]
=> P(A ∩ D) = P(A ∩ B) + P(A ∩ C) - P[(A ∩ B) ∩ (A ∩ C)]
=> P(A ∩ D) = P(A ∩ B) + P(A ∩ C) - P(A ∩ B ∩ C) .................2
Again, P(D) = P(B ∪ C) = P(B) + P(C) - P(B ∩ C) ....................3
Now, from equation 1, we get
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(B ∩ C) - [P(A ∩ B) + P(A ∩ C) - P(A ∩ B ∩ C)]
=> P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(B ∩ C) - P(A ∩ B) - P(A ∩ C) + P(A ∩ B ∩ C)
=> P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(B ∩ C) - P(A ∩ C) + P(A ∩ B ∩ C)
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