Math Problem Statement

For the experiment of rolling an ordinary pair of dice, find the probability that the sum will be even or a multiple of 6.

Solution

Problem Analysis

The problem asks to find the probability that the sum of the numbers obtained from rolling two ordinary dice is either:

  1. Even, or
  2. A multiple of 6.

Step-by-Step Solution

  1. Total Outcomes: When rolling a pair of dice, each die can show a number from 1 to 6. Therefore, there are a total of 6×6=366 \times 6 = 36 equally likely outcomes.

  2. Finding Sums that Satisfy the Conditions: Let's identify the sums that meet the given conditions.

    • Condition 1: Even Sum: The possible even sums from rolling two dice are 2, 4, 6, 8, 10, and 12. We count the number of combinations that result in these sums:

      • Sum = 2: (1, 1)
      • Sum = 4: (1, 3), (2, 2), (3, 1)
      • Sum = 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
      • Sum = 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
      • Sum = 10: (4, 6), (5, 5), (6, 4)
      • Sum = 12: (6, 6)

      Total combinations for even sums: 1+3+5+5+3+1=181 + 3 + 5 + 5 + 3 + 1 = 18.

    • Condition 2: Multiple of 6: The multiples of 6 from rolling two dice are 6 and 12. We count the number of combinations for these sums:

      • Sum = 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
      • Sum = 12: (6, 6)

      Total combinations for multiples of 6: 5+1=65 + 1 = 6.

  3. Combining the Two Conditions Using the Inclusion-Exclusion Principle: We need the probability that the sum is either even or a multiple of 6. The overlap between these two sets is the number of sums that are both even and multiples of 6, which are the sums 6 and 12.

    • Overlap count = 6 (as calculated above).

    Using the Inclusion-Exclusion principle: Total Count=(Even Sum Count)+(Multiple of 6 Count)(Overlap Count)\text{Total Count} = (\text{Even Sum Count}) + (\text{Multiple of 6 Count}) - (\text{Overlap Count}) Total Count=18+66=18\text{Total Count} = 18 + 6 - 6 = 18

  4. Probability Calculation: The probability is given by: Total CountTotal Outcomes=1836=12\frac{\text{Total Count}}{\text{Total Outcomes}} = \frac{18}{36} = \frac{1}{2}

Final Answer

The probability that the sum of the pair of dice is either even or a multiple of 6 is: 12\boxed{\frac{1}{2}}

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Related Questions:

  1. What is the probability that the sum of two dice is a prime number?
  2. How many outcomes result in an odd sum?
  3. If three dice are rolled, what is the probability of the sum being a multiple of 3?
  4. What is the probability that the sum is greater than 8?
  5. How does the probability change if one die is biased to show only even numbers?

Tip:

To handle multiple conditions in probability problems, always check for overlaps using the Inclusion-Exclusion principle.

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Math Problem Analysis

Mathematical Concepts

Probability
Combinatorics

Formulas

Probability of event = (Favorable outcomes) / (Total outcomes)

Theorems

Inclusion-Exclusion Principle

Suitable Grade Level

Grades 7-10