Ajith engages in a game where he simultaneously tosses

Solution

When 3 coins are tossed at once, set of all possible events = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} = 8 events So, probability of obtaining 3 heads = (1/8) When two die at thrown at once, set of all possible events = {(1, 1), (1, 2), (1, 3) .. (6, 6)} = 36 events Number of events where the sum obtained is at least 10 = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)} = 6 events So, probability of obtaining a sum greater than or equal to 10 = (6/36) = (1/6) Therefore, probability of Ajith winning the game = (1/8) X (1/6) = (1/48)