Two fair dice are thrown. Find the probability of getting the sum of numbers appeared a multiple of 3 or a multiple of 4.

Total possible outcomes when two fair dice are thrown = 6 × 6 = 36. We want: P(sum is a multiple of 3 or a multiple of 4). Sums that are multiples of 3: 3, 6, 9, 12 Count outcomes: Sum = 3: (1,2), (2,1) → 2 Sum = 6: (1,5), (2,4), (3,3), (4,2), (5,1) → 5 Sum = 9: (3,6), (4,5), (5,4), (6,3) → 4 Sum = 12: (6,6) → 1 Total multiples of 3 = 2 + 5 + 4 + 1 = 12 Sums that are multiples of 4: 4, 8, 12 Count outcomes: Sum = 4: (1,3), (2,2), (3,1) → 3 Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 Sum = 12: (6,6) → 1 Total multiples of 4 = 3 + 5 + 1 = 9 Sum = 12 is counted in both sets, so subtract intersection once: Number of favorable outcomes = 12 + 9 − 1 = 20 Probability = 20 / 36 = 5 / 9