When tossing two six-sided dice, there are $6 \times 6 = 36$ possible outcomes because each die can show one of 6 faces. Let's now calculate the theoretical probabilities for each case:

A. Probability that both dice show the same number:

The only way the two dice show the same number is when they both show $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$. There are 6 such outcomes.

$P(\text{both dice show the same number}) = \frac{6}{36} = \frac{1}{6}$

B. Probability that both dice show odd numbers:

The odd numbers on a die are 1, 3, and 5. For both dice to show odd numbers, each die must show either 1, 3, or 5. So, there are $3 \times 3 = 9$ favorable outcomes.

$P(\text{both dice show odd numbers}) = \frac{9}{36} = \frac{1}{4}$

C. Probability that the sum of the two numbers is 8:

To get a sum of 8, the following combinations are possible: $(2,6), (3,5), (4,4), (5,3), (6,2)$ There are 5 favorable outcomes.

$P(\text{sum of the two numbers is 8}) = \frac{5}{36}$

D. Probability that the sum of the two numbers is 10:

To get a sum of 10, the following combinations are possible: $(4,6), (5,5), (6,4)$ There are 3 favorable outcomes.

$P(\text{sum of the two numbers is 10}) = \frac{3}{36} = \frac{1}{12}$

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Let me know if you'd like more details on any of these results!

Expanding questions:

1. What is the probability that the sum of the two dice is less than 5?
2. What is the probability that the sum of the two dice is 7?
3. What is the probability that both dice show different even numbers?
4. What is the probability that the product of the two dice is an even number?
5. What is the probability that one die shows a prime number?

Tip: When dealing with probabilities for dice rolls, it's useful to list out all possible outcomes (like in a table) to systematically find the desired events.