When a dice is thrown twice, the all possible outcomes are
(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
Number of the outcomes which contain digit 5 = 11.
The number which does not contain the digit 5
= 36 – 11 = 25
(i) Number of favourable outcomes not occurring 5 = 25
Total number of all possible outcomes = 36
The required probability
= \(\frac { 25 }{ 36 }\)
(ii) The number of favourable outcomes of occurring a digit 5 at least = 11
Total number of all possible outcomes = 36
∴ The probability that the digit 5 occurs at least one time = \(\frac { 11 }{ 36 }\)