Question:

The total number of possible outcomes when a die is rolled three times is $6 \times 6 \times 6 = 6^3 = 216$. For a favorable outcome, the number rolled each time must be strictly larger than the previous number. Let the outcomes of the three rolls be $a, b, c$. The condition is $a < b < c$. This requires selecting 3 distinct numbers from the set of possible die rolls $\{1, 2, 3, 4, 5, 6\}$. Once 3 distinct numbers are chosen, there is only one way to arrange them in strictly increasing order. The number of ways to choose 3 distinct numbers from 6 is given by the combination formula: $$ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 $$ Thus, there are 20 favorable outcomes. The probability of getting a larger number than the previous number each time is the ratio of favorable outcomes to the total number of outcomes: $$ P = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{20}{216} $$ Simplifying the fraction: $$ P = \frac{20 \div 4}{216 \div 4} = \frac{5}{54} $$