For 3 sets A , B & C

n(A) = Number of elements of set A

n(B) = Number of elements of set B

n(C) = Number of elements of set C

 

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)

 

Proof of n(A ∪ B ∪ C) Formula

We know that

P(E ∪ F) = P(E) + P(F) − P(E ∩ F).

 

Putting E = B, F = (B ∪ C)

P(A ∪ (B ∪ C)) = P(A) + P(B ∪ C) − P(A ∩ (B ∪ C)),

 

And also

P(A ∩ (B ∪ C)) = P((A ∩ B) ∪ (A ∩ C))

  = P(A ∩ B) + P(A ∩ C) − P(A ∩ B ∩ A ∩ C)

  = P(A ∩ B) + P(A ∩ C) − P(A ∩ B ∩ C)

and

P(B ∪ C) = P(B) + P(C) − P(B ∩ C).

 

Substituting the 2 nd and 3 rd equations into the first 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)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.