When a coin is tossed twice, the digits on the top faces may be
(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).
∴ Total number of all possible outcomes = 36
(i) The favourable outcomes of getting the sum 9 = (6, 3), (5, 4), (4, 5), (3, 6)
The number of favourable outcomes = 4
∴ Required probability = \(\frac { 4 }{ 36 }\) = \(\frac { 1 }{ 9 }\)
(ii) When the dice is thrown two times, then the favourable outcomes of getting a sum 13 = 0.
Hence the probability = 0.
