Two unbiased dice are rolled and the numbers obtained are added. If the probability of getting the sum S is P(S), which of the following statements are true?

P(10) + P(11) = P(6)
P(6) > P(8)
P(2) + P(3) + P(4) > P(7)
P(7) = 1.5(P(5))
P(11) = P(3)

Let's analyze the probabilities related to the sums when rolling two unbiased dice. First, we need to calculate the probabilities for each possible sum \( S \) from 2 to 12. The sums and their probabilities are based on the possible combinations that result in each sum.

### Probabilities for sums:

- **Sum = 2:** (1,1) → 1 way → \( P(2) = \frac{1}{36} \)
- **Sum = 3:** (1,2), (2,1) → 2 ways → \( P(3) = \frac{2}{36} = \frac{1}{18} \)
- **Sum = 4:** (1,3), (2,2), (3,1) → 3 ways → \( P(4) = \frac{3}{36} = \frac{1}{12} \)
- **Sum = 5:** (1,4), (2,3), (3,2), (4,1) → 4 ways → \( P(5) = \frac{4}{36} = \frac{1}{9} \)
- **Sum = 6:** (1,5), (2,4), (3,3), (4,2), (5,1) → 5 ways → \( P(6) = \frac{5}{36} \)
- **Sum = 7:** (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 ways → \( P(7) = \frac{6}{36} = \frac{1}{6} \)
- **Sum = 8:** (2,6), (3,5), (4,4), (5,3), (6,2) → 5 ways → \( P(8) = \frac{5}{36} \)
- **Sum = 9:** (3,6), (4,5), (5,4), (6,3) → 4 ways → \( P(9) = \frac{4}{36} = \frac{1}{9} \)
- **Sum = 10:** (4,6), (5,5), (6,4) → 3 ways → \( P(10) = \frac{3}{36} = \frac{1}{12} \)
- **Sum = 11:** (5,6), (6,5) → 2 ways → \( P(11) = \frac{2}{36} = \frac{1}{18} \)
- **Sum = 12:** (6,6) → 1 way → \( P(12) = \frac{1}{36} \)

Now, let's verify the statements:

1. **Statement: \( P(10) + P(11) = P(6) \)**
   - Calculating:
     - \( P(10) + P(11) = \frac{1}{12} + \frac{1}{18} = \frac{3}{36} + \frac{2}{36} = \frac{5}{36} \)
     - \( P(6) = \frac{5}{36} \)
   - **True**.

2. **Statement: \( P(6) > P(8) \)**
   - \( P(6) = \frac{5}{36} \)
   - \( P(8) = \frac{5}{36} \)
   - **False**, since \( P(6) = P(8) \).

3. **Statement: \( P(2) + P(3) + P(4) > P(7) \)**
   - \( P(2) + P(3) + P(4) = \frac{1}{36} + \frac{2}{36} + \frac{3}{36} = \frac{6}{36} = \frac{1}{6} \)
   - \( P(7) = \frac{6}{36} = \frac{1}{6} \)
   - **False**, since \( P(2) + P(3) + P(4) = P(7) \).

4. **Statement: \( P(7) = 1.5 \cdot P(5) \)**
   - \( P(7) = \frac{1}{6} \)
   - \( 1.5 \cdot P(5) = 1.5 \cdot \frac{1}{9} = \frac{3}{18} = \frac{1}{6} \)
   - **True**.

5. **Statement: \( P(11) = P(3) \)**
   - \( P(11) = \frac{2}{36} = \frac{1}{18} \)
   - \( P(3) = \frac{2}{36} = \frac{1}{18} \)
   - **True**.

### Summary:
- **True Statements:** 1, 4, 5.
- **False Statements:** 2, 3.

Would you like more details, or do you have any questions? 

Here are some related questions to explore further:
1. How does the probability distribution of the sums of two dice compare to other dice games?
2. What is the expected value of the sum when rolling two dice?
3. How would the probabilities change if the dice were biased?
4. Can you calculate the probability of getting an even sum?
5. What are the probabilities of sums when rolling three dice instead of two?

**Tip:** Understanding probability distributions is crucial in games of chance and can help in strategic decision-making!

Explore the probabilities of different sums when rolling two unbiased dice. This problem checks the truth of several probability statements, such as P(10) + P(11) = P(6) and P(7) = 1.5(P(5)), by using combinatorial probability concepts. Suitable for Grades 8-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation