Let's solve this step by step:

1. Total possible outcomes (without restrictions)

When two dice are thrown, each die has 6 faces, so the total number of possible outcomes is: $6 \times 6 = 36$ These are all the combinations of two dice.

2. Restricting the outcomes (different numbers)

Since it's given that the numbers on the two dice are different, we eliminate the outcomes where both numbers are the same. These identical outcomes are: $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$ There are 6 such outcomes. Thus, the number of outcomes where the numbers are different is: $36 - 6 = 30$

3. Favorable outcomes (sum is 4)

Next, we find the pairs of different numbers on the dice that sum to 4. These pairs are: $(1, 3), (3, 1), (2, 2)$ However, we are told that the numbers must be different, so we exclude $(2, 2)$. The remaining favorable outcomes are: $(1, 3) \quad \text{and} \quad (3, 1)$ There are 2 favorable outcomes.

4. Calculating the probability

The probability is the ratio of favorable outcomes to the total number of outcomes with different numbers. Therefore, the probability is: $P = \frac{2}{30} = \frac{1}{15}$

Final Answer:

The probability that the sum of the two numbers is 4, given that the numbers are different, is: $\boxed{\frac{1}{15}}$

Would you like more details on this or have any questions?

Related Questions:

1. What is the probability that the sum is 5, given that the numbers are different?
2. How would the probability change if both numbers were allowed to be the same?
3. What is the probability that the sum is less than 6, given that the numbers are different?
4. If two dice are thrown, what is the probability that their sum is 8 and the numbers are different?
5. What is the probability that the two dice show a prime sum, given that the numbers are different?

Tip:

When calculating probabilities with dice, always first clarify if there are any restrictions (e.g., distinct numbers, sum conditions) to avoid errors.