Subscribe to our Youtube Channel - https://you.tube/teachoo
Ex 16.3, 16 - Chapter 16 Class 11 Probability
Last updated at Feb. 11, 2020 by Teachoo
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 (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
Ex 16.3
Ex 16.3, 1
Ex 16.3, 2
Ex 16.3, 3 Important
Ex 16.3, 4 Important
Ex 16.3 ,5 Important
Ex 16.3, 6
Ex 16.3, 7 Important
Ex 16.3, 8 Important
Ex 16.3, 9
Ex 16.3, 10
Ex 16.3, 11 Important
Ex 16.3, 12 Important
Ex 16.3, 13
Ex 16.3, 14 Important
Ex 16.3, 15 Important
Ex 16.3, 16 Important You are here
Ex 16.3, 17
Ex 16.3, 18
Ex 16.3, 19
Ex 16.3, 20 Important
Ex 16.3, 21 Important
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.