A fair die has six faces numbered 1, 1, 1, 2, 2, and 3. The die is rolled twice and the number showing on the uppermost face is recorded. Find the probability that the sum of the two numbers is at least three.

There are a couple of ways to approach this. A grid showing all the possibilities can be drawn up, which would help to illustrate the solution a little more explicitly. For now, let's talk through the logic. The event we are interested in is that the sum of the two numbers is at least three. In algebra, if we let S be the sum of the two rolls, then we require S ⩾ 3. So any sum greater than three fulfils this criteria as well as if the sum is exactly three. The wording "more than three" would imply the inequality S > 3, which would give a different answer. What possible sums can we obtain? The lowest would be rolling a "1" twice - in which case, we would get 1 + 1 = 2, which is strictly less than three and is not what we're after. The next lowest would be obtaining a "1" and a "2" in either order. The sum would be 1 + 2 = 3, which IS at LEAST as big as three. All the other combinations give sums greater than three, which also satisfy the criterion.
Next, we have to count how many ways there are to obtain each sum. As mentioned before, a grid can be used to show this more visually. If we think about how many ways there are to obtain a sum smaller than three, we have to roll two "1"s. On each roll, there are three "1"s that the die could land on. Hence, because the two rolls have no effect on each other (the technical term is independent), we multiply these combinations to give us 3 x 3 = 9 possible ways to roll two "1"s. Using the same logic, there are six possible options on each roll, so with two rolls we have 6 x 6 = 36 possible combinations. Now, nine combinations give a sum of less than three, so the remaining 36 - 9 = 27 combinations have a sum at least as big as three. So the probability of obtaining a sum at least as big as three is 27/36, or simplifying, 3/4 (or 0.75). There are alternative, more algebraic solutions too.