Examples

Chapter 16 Class 11 Probability
Serial order wise

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Example 16 If A, B, C are three events associated with a random experiment, prove that P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) – P (B ∩ C) + P ( A ∩ B ∩ C) Let B ∪ C = E So, P (A ∪ B ∪ C) = P(A ∪ E) = P(A) + P(E) – P(A ∩ E) = P(A) + P(E) – P(A ∩ (B ∪ C)) = P(A) + P(E) – P((A ∩ B) ∪ (A ∩ C)) We find P(E) & P((A ∩ B) ∪ (A ∩ C)) separately Finding P(E) P (E) = P (B ∪ C) = P(B) + P(C) − P(B ∩ C) Finding P((A ∩ B) ∪ (A ∩ C)) We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Putting A = (A ∩ B) & B = (A ∩ C) P((A ∩ B) ∪ (A ∩ C)) = P(A ∩ B) + P(A ∩ C) – P ((A ∩ B) ∩ (A ∩ C)) P((A ∩ B) ∪ (A ∩ C)) = P(A ∩ B) + P(A ∩ C) – P (A ∩ B ∩ C) Putting (2) & (3) in (1) P (A ∪ B ∪ C) = P(A) + P(E) – P((A ∩ B) ∪ (A ∩ C)) = P(A) + [P(B) + P(C) – P(B ∩ C)] – [P(A ∩ B) + P(A ∩ C) + P(A ∩ B ∩ C)] = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) – P (B ∩ C) + P ( A ∩ B ∩ C) Hence proved

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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.