Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Example 14 - Chapter 13 Class 12 Probability (Term 2)
Last updated at Feb. 15, 2020 by Teachoo
Examples
Example 1
Example 2
Example 3
Example 4
Example 5 Important
Example 6
Example 7 Important
Example 8
Example 9 Important
Example 10
Example 11 Important
Example 12 Important
Example 13 Important
Example 14 Important You are here
Example 15 Important
Example 16
Example 17 Important
Example 18 Important
Example 19 Important
Example 20 Important
Example 21 Important
Example 22
Example 23
Example 24 Important
Example 25 Important
Example 26 Important
Example 27
Example 28 Important Deleted for CBSE Board 2023 Exams
Example 29 Important
Example 30 Important Deleted for CBSE Board 2023 Exams
Example 31 Important Deleted for CBSE Board 2023 Exams
Example 32 Important Deleted for CBSE Board 2023 Exams
Example 33 Important
Example 34 Deleted for CBSE Board 2023 Exams
Example 35
Example 36 Important
Example 37 Important
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′) Two events A and B are independent if P(A ∩ B) = P(A) . P(B) 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 So, P(at least one of A & B) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = P(A) + P(B) – P(A ∩ 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’) ∴ P(at least one of A & B) = 1 – P(A’) P(B’) Hence Proved Given, A & B are independent So, P(A ∩ B) = P(A) . P(B)
