Two fair dice are tossed together once. What is the probability of getting a total of at least 9 from the outcome?

Analyze the question: Two fair dice are tossed, and we want P(sum ≥ 9). The total number of outcomes is 6 × 6 = 36.

Concept: Probability = (favourable outcomes) ÷ (total outcomes). Count the ways to get each total from 9 to 12.

Solution:

• Sum = 9: (3,6), (4,5), (5,4), (6,3), which is 4 ways
• Sum = 10: (4,6), (5,5), (6,4), which is 3 ways
• Sum = 11: (5,6), (6,5), which is 2 ways
• Sum = 12: (6,6), which is 1 way

Total favourable = 4 + 3 + 2 + 1 = 10

P(sum ≥ 9) = 10/36 = 5/18

Answer: Option C, 5/18.

Tip: For dice sums, the number of ways rises by 1 up to 7 and then falls by 1. The counts for sums 2 to 12 are 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1. For "at least 9", add the last four: 4+3+2+1 = 10. Always simplify the fraction over 36.