To solve this problem, let's first consider the total number of possible outcomes when two dice are thrown.

Total Possible Outcomes:

Each die has 6 faces, so when two dice are thrown, the total number of possible outcomes is:

$6 \times 6 = 36$

Favorable Outcomes:

Now, we need to find the probability that at least one die shows a number greater than 3. The numbers greater than 3 on a die are 4, 5, and 6. We can count the outcomes where at least one die shows a 4, 5, or 6.

Case 1: First die shows 4, 5, or 6:

• If the first die shows 4, 5, or 6, the second die can show any of the 6 numbers (1, 2, 3, 4, 5, or 6).
• So, there are $3 \times 6 = 18$ favorable outcomes.

Case 2: Second die shows 4, 5, or 6 (and the first die shows a number less than or equal to 3):

• If the second die shows 4, 5, or 6 and the first die shows 1, 2, or 3.
• So, there are $3 \times 3 = 9$ additional favorable outcomes.

Adding these gives $18 + 9 = 27$ favorable outcomes.

Probability:

The probability that at least one die shows a number greater than 3 is:

$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{27}{36} = \frac{3}{4}$

List of Favorable Outcomes:

The favorable outcomes where at least one die shows a number greater than 3 are:

[ (4,1), (4,2), (4,