Examples

Chapter 13 Class 12 Probability
Serial order wise

### Transcript

Example 14 If A and B are two independent events, then the probability of occurrence of at least one of A and B is given by 1– P(A′) P(B′) Since events A and B are independent ∴ P(A ∩ B) = P(A) . P(B) Now, Probability of occurrence of at least one of A and B = Probability of occurrence of only A + Probability of occurrence of only B + Probability of occurrence of A and B = Probability of occurrence of A or B = P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Since A & B are independent Putting P(A ∩ B) = P(A) . P(B) = P(A) + P(B) – P(A) . P(B) = P(A) + P(B) [ 1 – P(A) ] = P(A) + P(B) . P(A’) = 1 – P(A’) + P(B) . P(A’) = 1 – P(A’) [ 1 – P(B) ] = 1 – P(A’) . P(B’) Thus, P(at least one of A & B) = 1 – P(A’) P(B’) Hence Proved

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.