Ajith engages in a game where he simultaneously tosses three coins and rolls a pair of dice. To win, he must achieve two outcomes: all three coins must land as heads, and the sum of the numbers shown on the two dice must be at least 10. What is the probability of Ajith winning the game?

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)