Two dice are rolled. Determine the probability that the number on the first die's upper face is less than the number on the second die's upper face:

The sample space for the above experiment can be seen below:

{ (1,1) (2,1) (3,1) (4,1) (5,1) (6,1)

​(1,2) (2,2) (3,2) (4,2) (5,2) (6,2)

​(1,3) (2,3) (3,3) (4,3) (5,3) (6,3)

​(1,4) (2,4) (3,4) (4,4) (5,4) (6,4)

​(1,5) (2,5) (3,5) (4,5) (5,5) (6,5)

​(1,6) (2,6) (3,6) (4,6) (5,6) (6,6)​ }
 

To calculate the probability of each event, we can use the fact that the probability of any event is the ratio of the number of favourable outcomes to the total number of possible outcomes.

Total number of possible outcomes $=6\times 6=36$  (As the two dice are independent)

From the sample space, number of times the number obtained on the first dice is less than second $=15$

$\mathrm{Probability} = \frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{possible} \mathrm{outcomes} }$

Therefore, probability  $=\frac{15}{36}$

                                          $=\frac{5}{12}$

Hence, the correct answer is $\frac{5}{12}$.