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.