Example 14 - Chapter 13 Class 12 Probability
Last updated at April 16, 2024 by Teachoo
Independent events
Ex 13.2, 6
Ex 13.2, 10 Important
Ex 13.2, 5
Example 10
Example 11 Important
Example 12 Important
Ex 13.2, 15 (i)
Ex 13.2, 8
Ex 13.2, 7 Important
Ex 13.2, 11 (i)
Ex 13.2, 4
Ex 13.2, 13 Important
Ex 13.2, 14 Important
Ex 13.2, 18 (MCQ) Important
Example 13 Important
Example 14 Important You are here
Independent events
Last updated at April 16, 2024 by Teachoo
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