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?

Step-by-step solution

Total number of possible outcomes when two fair dice are tossed is 6 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 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) = 6 36 = 1 6 . 2 P(S > 7) + 1 6 = 1 2 P(S > 7) = 1 - 1 6 = 5 6 P(S > 7) = 5 12 Answer: 5 12