For two dice or two types of trials, it is easier to organize the outcomes by writing a table (two-dimensional table).
Example ①
When two dice are rolled, find the probability that the sum of the numbers is 7.
Solution: In all, 6×6=36 outcomes. The combinations whose sum is 7 are (1,6)(2,5)(3,4)(4,3)(5,2)(6,1) → 6 outcomes. P = 6/36 = 1/6
When two dice are rolled, find the probability that the sum of the numbers is 7.
Solution: In all, 6×6=36 outcomes. The combinations whose sum is 7 are (1,6)(2,5)(3,4)(4,3)(5,2)(6,1) → 6 outcomes. P = 6/36 = 1/6
Example ②
When two dice are rolled, find the probability that at least one die shows 4 or more.
Solution: The complementary event, “both dice show 3 or less,” has 3×3=9 outcomes. The required probability = 1−9/36 = 1−1/4 = 3/4
When two dice are rolled, find the probability that at least one die shows 4 or more.
Solution: The complementary event, “both dice show 3 or less,” has 3×3=9 outcomes. The required probability = 1−9/36 = 1−1/4 = 3/4
Key Points
- Two dice → 6×6=36 outcomes (organized in a table)
- Count the cells that satisfy the condition
- For “at least,” the complementary event is efficient
Practice Problems
- When two dice are rolled, find the probability that the product of the numbers is 6.
Answers and Explanations
- Answer: 1/9. A product of 6 occurs in (1,6)(2,3)(3,2)(6,1), or 4 outcomes. 4/36=1/9.