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

Ex 16.3, 16 - Chapter 16 Class 11 Probability - Part 2

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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 (By Demorgan law) Demorgan’s law █("If (A’" ∩"B’) = (A " ∪" B)’ " @"or (A’ " ∪" B’) = (A " ∩" B)’" ) It means there is common elements between E and F Hence E and F are not mutually exclusive

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.