Math Problem Statement

Two dice are rolled. Determine the probability of the following. Rolling an even number or a number greater than 8

Solution

To solve this problem, we calculate the probability of rolling an even number or a number greater than 8 when rolling two dice.

Step 1: Total number of outcomes

When two dice are rolled, the total number of outcomes is: 6×6=366 \times 6 = 36

Step 2: Outcomes for each event

  • Rolling an even number: An even number on a die is 2,4,62, 4, 6. If one die shows an even number, there are 3×6=183 \times 6 = 18 favorable outcomes for one die being even.

  • Rolling a number greater than 8: The possible sums greater than 8 are 9,10,11,129, 10, 11, 12. Let’s calculate the favorable pairs for each sum:

    • Sum = 99: (3,6),(4,5),(5,4),(6,3)(3, 6), (4, 5), (5, 4), (6, 3)44 outcomes
    • Sum = 1010: (4,6),(5,5),(6,4)(4, 6), (5, 5), (6, 4)33 outcomes
    • Sum = 1111: (5,6),(6,5)(5, 6), (6, 5)22 outcomes
    • Sum = 1212: (6,6)(6, 6)11 outcome
      Total = 4+3+2+1=104 + 3 + 2 + 1 = 10 outcomes

Step 3: Overlapping outcomes

We need to account for outcomes that satisfy both conditions (rolling an even number and a sum greater than 8):

  • Sum 1010: (4,6),(6,4)(4, 6), (6, 4)22 even outcomes
  • Sum 1212: (6,6)(6, 6)11 even outcome
    Total overlap = 2+1=32 + 1 = 3 outcomes

Step 4: Applying the inclusion-exclusion principle

The formula for the probability of ABA \cup B is: P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B) Using the counts:

  • P(A)=1836=0.5P(A) = \frac{18}{36} = 0.5
  • P(B)=1036=0.2778P(B) = \frac{10}{36} = 0.2778
  • P(AB)=336=0.0833P(A \cap B) = \frac{3}{36} = 0.0833

P(AB)=0.5+0.27780.0833=0.6944P(A \cup B) = 0.5 + 0.2778 - 0.0833 = 0.6944

Final Probability

The probability of rolling an even number or a number greater than 8 is: 0.6944or approximately 69.44%\boxed{0.6944 \, \text{or approximately } 69.44\%}


Would you like more details or clarifications? Here are some related questions:

  1. What is the probability of rolling a sum exactly equal to 8 with two dice?
  2. How many outcomes result in rolling a sum less than 6 with two dice?
  3. Can we extend this calculation to three dice instead of two?
  4. How do probabilities change if we roll biased dice instead of fair dice?
  5. What are the odds of rolling doubles when two dice are thrown?

Tip: Using Venn diagrams can simplify calculations involving overlapping probabilities!

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Math Problem Analysis

Mathematical Concepts

Probability
Set Theory
Inclusion-Exclusion Principle

Formulas

Total outcomes = number of sides on die 1 × number of sides on die 2
Probability of event = favorable outcomes / total outcomes
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

Theorems

Inclusion-Exclusion Principle

Suitable Grade Level

Grades 9-11