Total number of possible outcomes = 36
(i) Outcomes favourable to given event are: (4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)
Number of favourable outcomes = 6
P (Sum atleast ten) = 6/36=1/6
(ii) Outcomes favourable to given event are: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
Number of favourable outcomes = 6
P(a doubtlet) =6/36=1/6
(iii) Outcomes favourable to given event are: (1, 6), (2, 3), (3, 2), (6, 1)
Number of favourable outcomes = 4
P (six as a product) =4/36=1/9