Ex 14.1, 5 (ii) - Chapter 14 Class 11 Probability
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Ex 14.1, 5 Three coins are tossed. Describe (ii) Three events which are mutually exclusive and exhaustive. S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} Let A be the event getting exactly two tail comes A = {HTT, THT, TTH} Let B be the event getting at least two head B = {HHT, HTH, THH, HHH} Let C be the event getting only tail C = {TTT} A ∩ B = {HTT, THT, TTH} ∩ {HHT, HTH, THH, HHH} = 𝜙 Since no element is common in A & B Hence, A & B are mutually exclusive B ∩ C = {HHT, HTH, THH, HHH} ∩ {TTT} = 𝜙 Since no element is common in B & C Hence, B & C are mutually exclusive A ∩ C = {HHT, THT, TTH} ∩ {TTT} = 𝜙 Since no element is common in A & C Hence, A & C are mutually exclusive Since A & B, A & C, B & C are mutually exclusive Hence A, B and C are mutually exclusive Also, A ∪ B ∪ C = {HTT, THT, TTH, HHT, HTH, THH, HHH} = S Hence A, B & C are exhaustive events
About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo