Ex 16.3, 16 - P(not E or not F) = 0.25, State whether E, F - Using formulae of sets

Slide51.JPG

  1. Chapter 16 Class 11 Probability
  2. Serial order wise
Ask Download

Transcript

Ex 16.3, 16 Events E and F are such that P(not E or not F) = 0.25, State whether E and F are mutually exclusive. Given that P (not E or not F) = 0.25 P (E’ ∪ F’) = 0.25 P (E ∩ F)’ = 0.25 1 – P (E ∩ F) = 0.25 1 – 0.25 = P (E ∩ F) 0.75 = P (E ∩ F) P (E ∩ F) = 0.75 Since P (E ∩ F) ≠ 0 Since P (E ∩ F) ≠ 0 It means there is common elements between E and F Hence E and F are not mutually exclusive

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.