A die is rolled thrice, the probability of getting a larger number each time than the previous number, is

$\frac{5}{54}$

We have,

Total number of elementary events = 6 × 6 × 6=216.

Clearly, the second number has to be greater than unity. If the second number is i (i>1), then the first can be chosen in (i-1) ways and the third (6–1) ways. So, three numbers can be chosen in (i-1) ×1 (6-1) ways. But, the second number can vary from 2 to 5. Therefore,
Favourable number of elementary events

$= \sum\limits^{5}_{i=2}(i-1)(6-i) = 1 × 4 +2 ×3 +3 × 2 + 4 × 1 = 20 $

Hence, required probability $=\frac{20}{216}=\frac{5}{54}$