


Ex 16.3
Ex 16.3, 2
Ex 16.3, 3 (i) Important
Ex 16.3, 3 (ii)
Ex 16.3, 3 (iii)
Ex 16.3, 3 (iv)
Ex 16.3, 3 (v)
Ex 16.3, 4 Important
Ex 16.3 ,5 Important You are here
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 (i)
Ex 16.3, 12 (ii) Important
Ex 16.3, 13
Ex 16.3, 14 Important
Ex 16.3, 15 Important
Ex 16.3, 16 Important
Ex 16.3, 17
Ex 16.3, 18
Ex 16.3, 19
Ex 16.3, 20 Important
Ex 16.3, 21 Important
Ex 16.3
Last updated at Feb. 11, 2020 by Teachoo
Ex 16.3, 5 A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (i) 3 If the coin is tossed we get only 1 or 6 If a die is thrown we get 1, 2, 3, 4, 5, 6 Hence, S = {β("(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)," @" (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)" )} n(S) = 12 Let A be the event sum 3 Hence A = {(1, 2)} n(A) = 1 Probability of getting sum as 3 = P(A) = (n(A))/(n(S)) = π/ππ Ex 16.3, 5 A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (ii) 12 Let B be the event sum is 12. Hence B = {(6, 6)} n(B) = 1 Probability of getting sum as 12 = P(B) = (n(B))/(n(S)) = π/ππ