Two dices contain the numbers 0, 1, 1, 1, 6, 6 on the faces each.
The all possible outcomes will be:
(0, 0), (0, 1), (0, 1), (0, 1), (0, 6), (0, 6)
(1, 0), (1, 1), (1, 1), (1, 1), (1, 6), (1, 6)
(1, 0), (1, 1), (1, 1), (1, 1), (1, 6), (1, 6)
(1,0), (1, 1), (1, 1), (1, 1), (1, 6), (1, 6)
(6, 0), (6, 1), (6, 1), (6, 1), (6, 6), (6, 6)
(6, 0), (6, 1), (6, 1), (6, 1), (6, 6), (6, 6)
(i) The possible numbers which have different outcomes are (0, 0), (0, 1), (0, 6), (1, 0), (1, 1), (1, 6), (6, 6)
Hence, the possible outcomes are : (0, 1, 2, 6, 7, 12)
(ii) Total number of all possible outcomes = 36
The numbers that have a sum 7 are
(1,6), (1,6), (1,6), (1,6), (1,6), (1,6)
(6, 1), (6, 1), (6, 1), (6, 1), (6, 1), (6, 1)
The number of favourable outcomes = 12
∴ The probability of getting a sum of 7 = \(\frac { 12 }{ 36 }\) = \(\frac { 1 }{ 3 }\)