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

To 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 when two dice are rolled, we can use the following approach:

There are a total of 6 sides on each die, numbered from 1 to 6. We need to count the number of outcomes where the first die's number is less than the second die's number.

We can use a table to visualize the outcomes:

Now, count the number of favorable outcomes (where the first die's number is less than the second die's number):

For the first row, there are 5 favorable outcomes.
For the second row, there are 4 favorable outcomes.
For the third row, there are 3 favorable outcomes.
For the fourth row, there are 2 favorable outcomes.
For the fifth row, there is 1 favorable outcome.
For the sixth row, there are 0 favorable outcomes.
Now, sum up the favorable outcomes:

5 + 4 + 3 + 2 + 1 + 0 = 15

There are 15 favorable outcomes.

Next, calculate the total number of possible outcomes when two dice are rolled. Since each die has 6 sides, there are 6 x 6 = 36 possible outcomes.

Now, calculate the probability:

Probability = $\frac{ (\mathrm{Number} \mathrm{of} \mathrm{Favorable} \mathrm{Outcomes}) }{(\mathrm{Total} \mathrm{Number} \mathrm{of} \mathrm{Possible} \mathrm{Outcomes})}$
Probability = $\frac{15}{36}=\frac{5}{12}$

So, the probability that the number on the first die's upper face is less than the number on the second die's upper face when two dice are rolled is 5/12.