Two dice are thrown. What is the probability that the sum of the two dice is greater than $3$?
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Two dice are thrown. What is the probability that the sum of the two dice is greater than $3$?
Total outcomes = 36. Sum <= 3: (1,1), (1,2), (2,1) = 3 outcomes. Probability sum > 3 = 1 - 3/36 = 33/36 = 11/12.
The total number of outcomes when throwing two dice is 36, and we want the sum to be greater than 3. It is easier to find the probability of the complementary event, which is getting a sum of 2 or 3; the favorable combinations for this are (1,1), (1,2), and (2,1), totaling 3 outcomes. The probability of this complement is 3/36, or 1/12. Subtracting this from 1 gives 1 - 1/12 = 11/12.