If two "fair" dice are tossed what is the probability that the sum of their faces will be a multiple of 3? Show or explain how you got your answer?

possibilities

make a table
label 1 through 6 at top
label 1 through 6 down
you have 36 boxes
fill in the sums (1 through 12)
note symmetry and ease of filling in because every box is one plus the one to the left or above
a total of 12 of the 36 boxes holds a multiple of 3
12/36 = 1/3

To find the probability of getting a sum that is a multiple of 3 when two fair dice are tossed, we need to identify the favorable outcomes and the total number of possible outcomes.

Let's start by determining the total number of outcomes. When rolling two dice, each die has six possible outcomes (numbers 1 to 6). Since the two dice are rolled independently, the total number of outcomes will be 6 * 6 = 36.

Next, we need to find the favorable outcomes, which are the combinations of numbers that sum up to a multiple of 3. These combinations are:
(1, 2), (2, 1), (1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (4, 5), (5, 4), (5, 5), (6, 3), (3, 6)

Counting these combinations gives us 12 favorable outcomes.

Therefore, the probability of the sum of the two dice being a multiple of 3 is 12/36, which simplifies to 1/3.

To summarize:
Total number of outcomes (n) = 36
Number of favorable outcomes (x) = 12
Probability (P) = x/n = 12/36 = 1/3