Check sibling questions
Chapter 16 Class 11 Probability
Concept wise

Ex 16.2, 7 - Refer to question 6 above, State true or false

Ex 16.2, 7 - Chapter 16 Class 11 Probability - Part 2
Ex 16.2, 7 - Chapter 16 Class 11 Probability - Part 3
Ex 16.2, 7 - Chapter 16 Class 11 Probability - Part 4
Ex 16.2, 7 - Chapter 16 Class 11 Probability - Part 5
Ex 16.2, 7 - Chapter 16 Class 11 Probability - Part 6
Ex 16.2, 7 - Chapter 16 Class 11 Probability - Part 7
Ex 16.2, 7 - Chapter 16 Class 11 Probability - Part 8
Ex 16.2, 7 - Chapter 16 Class 11 Probability - Part 9
Ex 16.2, 7 - Chapter 16 Class 11 Probability - Part 10
Ex 16.2, 7 - Chapter 16 Class 11 Probability - Part 11
Ex 16.2, 7 - Chapter 16 Class 11 Probability - Part 12
Ex 16.2, 7 - Chapter 16 Class 11 Probability - Part 13

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Transcript

Ex 16.2, 7 Refer to question 6 above, State true or false: (give reason for your answer) A and B are mutually exclusive From 16.2 ,6 A = {█((2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6)"," @" " (4, 1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6)"," @" " (6, 1),(6, 2),(6, 3),(6, 4),(6, 5),(6 6) )} B = {█((1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(1, 6)"," @█((3, 1),(3, 2),(3, 3),(3, 4),(3, 5),(3, 6)@(5, 1),(5, 2),(5, 3),(5, 4),(5, 5),(5, 6) ))} A ∩ B = ϕ Since no common element in A & B So, A & B are mutually exclusive True. Ex 16.2, 7 Refer to question 6 above, State true or false: (give reason for your answer) (ii) A and B are mutually exclusive and exhaustive From 16.2 ,6 A = {█((2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6)"," @" " (4, 1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6)"," @" " (6, 1),(6, 2),(6, 3),(6, 4),(6, 5),(6 6) )} B = {█((1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(1, 6)"," @█((3, 1),(3, 2),(3, 3),(3, 4),(3, 5),(3, 6)@(5, 1),(5, 2),(5, 3),(5, 4),(5, 5),(5, 6) ))} A ∪ B = {█((1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(1, 6)@(2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6)@(3, 1),(3, 2),(3, 3),(3, 4),(3, 5),(3, 6)@(4, 1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6)@(5, 1),(5, 2),(5, 3),(5, 4),(5, 5),(5, 6)@(6, 1),(6, 2),(6, 3),(6, 4),(6, 5),(6 6),)}= S Since A ∪ B = S They are exhaustive events Also as per (i), they are mutually exclusive Hence they are mutually exclusive and exhaustive True Ex 16.2, 7 Refer to question 6 above, State true or false: (give reason for your answer) (iii) A = B’ A = getting even number on the first A = {█((2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6)"," @" " (4, 1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6)"," @" " (6, 1),(6, 2),(6, 3),(6, 4),(6, 5),(6 6) )} B = getting odd no on the first die = {█((1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(1, 6)"," @█((3, 1),(3, 2),(3, 3),(3, 4),(3, 5),(3, 6)@(5, 1),(5, 2),(5, 3),(5, 4),(5, 5),(5, 6) ))} B’ = getting even number on the first die = {█((2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6)"," @" " (4, 1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6)"," @" " (6, 1),(6, 2),(6, 3),(6, 4),(6, 5),(6 6) )} = A Hence A = B’ True. Ex 16.2, 7 Refer to question 6 above, State true or false: (give reason for your answer) (iv) A and C are mutually exclusive A = {█((2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6)"," @" " (4, 1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6)"," @" " (6, 1),(6, 2),(6, 3),(6, 4),(6, 5),(6 6) )} C = {█((1, 1), (1, 2), (1, 3), (1, 4),@(2, 1), (2, 2), (2, 3),@(3, 1), (3, 2), (4, 1))} A ∩ C = {(2, 1),(2, 2),(2, 3),(4, 1)} ≠ 𝜙 a Since there is common elements in A and C , So A & C are not mutually exclusive False. Ex 16.2, 7 Refer to question 6 above, State true or false: (give reason for your answer) (v) A and B’ are mutually exclusive A = {█((2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6)"," @" " (4, 1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6)"," @" " (6, 1),(6, 2),(6, 3),(6, 4),(6, 5),(6 6) )} B’ = {█((2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6)"," @" " (4, 1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6)"," @" " (6, 1),(6, 2),(6, 3),(6, 4),(6, 5),(6 6) )} A ∩ B’ = {█((2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6)"," @" " (4, 1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6)"," @" " (6, 1),(6, 2),(6, 3),(6, 4),(6, 5),(6 6) )} = A Since A ∩ B’ = A ≠ ϕ Hence there is common element between A & B’ Hence A and B’ is not mutually exclusive Hence, false Ex 16.2, 7 Refer to question 6 above, State true or false: (give reason for your answer) (vi) A,’ B’, C are mutually exclusive and exhaustive. A = getting an even number on the first die A’ = getting an odd number on the first die A’ = {█((1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(1, 6)"," @█((3, 1),(3, 2),(3, 3),(3, 4),(3, 5),(3, 6)@(5, 1),(5, 2),(5, 3),(5, 4),(5, 5),(5, 6) ))} B = getting an odd number on the first die B’ = getting an even number on the first die B’ = {█((2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6)"," @" " (4, 1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6)"," @" " (6, 1),(6, 2),(6, 3),(6, 4),(6, 5),(6 6) )} C = {█((1, 1), (1, 2), (1, 3), (1, 4),@(2, 1), (2, 2), (2, 3),@(3, 1), (3, 2), (4, 1))} A’ ∩ B’ = 𝝓 Hence there is no common element in A’ and B’ A & B are mutually exclusive B’ ∩ C = {(2, 1),(2, 2),(2, 3), (4, 1)} Hence there is common element between B’ and C Since B’ ∩ C ≠ 𝜙 Hence, B’ and C are not mutually exclusive Since B’ & C are not mutually exclusive A,’ B’, C are not mutually exclusive and exhaustive ∴ A,’ B’, C are not mutually exclusive and exhaustive Hence, False

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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.