Math Problem Statement

Two dice are rolled. Determine the probability of the following. ("Doubles" means both dice show the same number.) Rolling an even number or doubles

Solution

To solve this, we calculate the probability of rolling an even number or doubles using the principles of probability. Here's the step-by-step explanation:

Total Outcomes

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

Event 1: Rolling an Even Number

An even number can appear on either die. The even numbers are 2,4,62, 4, 6. For each die, there are 33 even outcomes. The number of outcomes where at least one die shows an even number is calculated using complementary counting (subtracting cases where both dice show odd numbers).

  • Both dice odd: Numbers 1,3,51, 3, 5, which means 33 choices for each die: 3×3=9 outcomes where both dice are odd.3 \times 3 = 9 \text{ outcomes where both dice are odd.}

  • At least one die even: 369=27 outcomes.36 - 9 = 27 \text{ outcomes.}

So, there are 2727 outcomes where at least one die shows an even number.

Event 2: Rolling Doubles

Doubles occur when both dice show the same number. The possible outcomes for doubles are: (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)(1,1), (2,2), (3,3), (4,4), (5,5), (6,6) This gives 66 outcomes.

Event 3: Intersection of Event 1 and Event 2

The intersection of "even number" and "doubles" occurs when the doubles are also even numbers. These outcomes are: (2,2),(4,4),(6,6)(2,2), (4,4), (6,6) This gives 33 outcomes.

Applying the Inclusion-Exclusion Principle

The probability of "even number or doubles" is calculated as: P(Even or Doubles)=P(Even)+P(Doubles)P(Even and Doubles)P(\text{Even or Doubles}) = P(\text{Even}) + P(\text{Doubles}) - P(\text{Even and Doubles})

Substitute the values: P(Even)=2736,P(Doubles)=636,P(Even and Doubles)=336P(\text{Even}) = \frac{27}{36}, \quad P(\text{Doubles}) = \frac{6}{36}, \quad P(\text{Even and Doubles}) = \frac{3}{36}

P(Even or Doubles)=2736+636336=3036=56P(\text{Even or Doubles}) = \frac{27}{36} + \frac{6}{36} - \frac{3}{36} = \frac{30}{36} = \frac{5}{6}

Final Answer:

The probability of rolling an even number or doubles is: 56\boxed{\frac{5}{6}}

Let me know if you would like a detailed breakdown or if you have questions!

Related Questions:

  1. What is the probability of rolling only doubles?
  2. What is the probability of rolling two even numbers?
  3. What is the probability of rolling an odd number on both dice?
  4. What is the probability of rolling a sum greater than 8?
  5. How does the probability change if one die is weighted?

Tip:

When calculating probabilities of combined events, always consider whether they overlap to avoid double-counting!

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Math Problem Analysis

Mathematical Concepts

Probability
Inclusion-Exclusion Principle
Combinatorics

Formulas

Total outcomes = 6 × 6 = 36
P(A or B) = P(A) + P(B) - P(A and B)

Theorems

Inclusion-Exclusion Principle

Suitable Grade Level

Grades 8-10