The problem asks for the probability that when rolling two identical cubic dice, at least one die shows a face value greater than 3.
A standard cubic die has faces numbered 1, 2, 3, 4, 5, 6.
It's often easier to calculate the probability of the event *not* happening and subtract it from 1.
The event we want is "at least one face value > 3".
The complement event is "neither face value is greater than 3", which means "both face values are less than or equal to 3".
The probability of the complement event (both dice $\le 3$) is:
$ P(\text{both} \le 3) = \frac{\text{Number of outcomes where both} \le 3}{\text{Total possible outcomes}} = \frac{9}{36} = \frac{1}{4} $Now, we can find the probability of the original event (at least one face value > 3) using the complement rule:
$ P(\text{at least one} > 3) = 1 - P(\text{both} \le 3) $ $ P(\text{at least one} > 3) = 1 - \frac{1}{4} $ $ P(\text{at least one} > 3) = \frac{3}{4} $The probability that at least one of the face values is greater than 3 is $\frac{3}{4}$.