A fair die is thrown two times. What is the probability that the sum of the scores is at least 10?

Analyze the question: We throw a fair die twice and want P(sum ≥ 10), meaning the sum is 10, 11, or 12.

Concept: Two throws give 6 × 6 = 36 equally likely outcomes. Probability = (favourable outcomes) ÷ (total outcomes).

Solution:

• Sum = 10: (4,6), (5,5), (6,4) → 3 outcomes
• Sum = 11: (5,6), (6,5) → 2 outcomes
• Sum = 12: (6,6) → 1 outcome

Favourable outcomes = 3 + 2 + 1 = 6

P(sum ≥ 10) = 6/36 = 1/6

Answer: The correct option is B, 1/6.

Tip: For two-dice questions, list the outcomes for each sum systematically, starting from the smallest sum that qualifies. The count for sums 10, 11, 12 follows the pattern 3, 2, 1. Also remember to simplify your fraction at the end, since 6/36 is not among the options until reduced.