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The problem asks: 

**"What is the probability that the sum of the numbers is not less than 10, when two dice are thrown simultaneously?"**

### Step-by-Step Solution:

1. **Identify possible sums when two dice are thrown:**
   The sums range from 2 (1+1) to 12 (6+6).

2. **Determine favorable outcomes (sum ≥ 10):**
   - To get a sum of 10, the pairs are: (4, 6), (5, 5), and (6, 4) → 3 outcomes.
   - To get a sum of 11, the pairs are: (5, 6), (6, 5) → 2 outcomes.
   - To get a sum of 12, the pair is: (6, 6) → 1 outcome.
   
   So, the total favorable outcomes for sum ≥ 10 are \( 3 + 2 + 1 = 6 \).

3. **Total possible outcomes:**
   Since each die has 6 faces, the total number of outcomes when two dice are thrown is:
   \[
   6 \times 6 = 36
   \]

4. **Calculate the probability:**
   The probability is the ratio of favorable outcomes to total outcomes:
   \[
   \frac{6}{36} = \frac{1}{6}
   \]

### Answer:
The probability that the sum is not less than 10 is **1/6**.

Let me know if you'd like more details or have any questions! Here are some follow-up questions to expand the topic:

1. What is the probability of getting a sum of exactly 7 when two dice are thrown?
2. How does the probability change if the dice are biased?
3. What is the probability of getting a sum greater than 8?
4. What are the odds in favor of getting a sum of 12?
5. How would the probability change if three dice were thrown instead of two?

**Tip:** The total number of possible outcomes when rolling two dice is always 36, which simplifies many probability calculations.

What is the probability that the sum of the numbers is not less than 10, when two dice are thrown simultaneously?

Learn how to calculate the probability that the sum of two dice is not less than 10. This problem involves basic probability theory and combinatorics, exploring favorable outcomes and total possibilities. Suitable for students in Grades 9-11.

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