24. A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once?

Outcomes are: (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) So, the total number of outcome = 6×6 = 36 (i) Method 1: Consider the following events. A = 5 comes in first throw, B = 5 comes in second throw P(A) = 6/36, P(B) = 6/36 and P(not B) = 5/6 So, P(not A) = 1-(6/36) = 5/6 ∴ The required probability = (5/6)×(5/6) = 25/36 Method 2: Let E be the event in which 5 does not come up either time. So, the favourable outcomes are [36–(5+6)] = 25 ∴ P(E) = 25/36 (ii) Number of events when 5 comes at least once = 11(5+6) ∴ The required probability = 11/36