A fair dice is tossed two times. The probability that the second toss results in a value that is higher than the first toss is

$\text{Total outcome are 36 out of which favorable outcomes are:} //// (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6); \\ (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6) \text{which are 15.} \ \text{Thus} \hspace{1cm} P(E) = \frac{\text{No. of favourable outcomes}}{\text{No. of total outcomes}}=\frac{15}{36} = \frac{5}{12}$