If two fair dice are tossed, then what is the probability that the...

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: $\dfrac{5}{12}$