Get live Maths 1-on-1 Classs - Class 6 to 12

Examples

Example 1
Deleted for CBSE Board 2023 Exams

Example 2 Important Deleted for CBSE Board 2023 Exams

Example 3 (i) Deleted for CBSE Board 2023 Exams

Example 3 (ii) Deleted for CBSE Board 2023 Exams

Example 4 Important Deleted for CBSE Board 2023 Exams

Example 5 Important Deleted for CBSE Board 2023 Exams

Example, 6

Example 7

Example, 8 Important

Example 9 Important

Example 10 Important

Example, 11

Example 12 Important

Example 13 Important

Example 14 Important

Example 15 Important

Example 16 You are here

Example 17 Important

Last updated at March 22, 2023 by Teachoo

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