If two fair dice are tossed, then what is the probability that the sum of the numbers on the faces of the dice is strictly greater than 7?

Total number of possible outcomes when two fair dice are tossed is $6 \times 6 = 36$.

Let $S$ be the sum of the numbers on the faces of the dice. By symmetry, the probability of getting a sum strictly greater than $7$ is equal to the probability of getting a sum strictly less than $7$.

$P(S > 7) + P(S < 7) + P(S = 7) = 1$

The outcomes for $S = 7$ are $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$, which are $6$ in number. Thus, $P(S = 7) = \dfrac{6}{36} = \dfrac{1}{6}$.

$2 P(S > 7) + \dfrac{1}{6} = 1$

$2 P(S > 7) = 1 - \dfrac{1}{6} = \dfrac{5}{6}$

$P(S > 7) = \dfrac{5}{12}$

Answer: $\dfrac{5}{12}$