Three fair six sided dice are thrown together. What is the probability that the total (sum of the three faces) is at least 6?

Introduction / Context:
This question deals with the distribution of sums obtained when three fair dice are rolled together. We must find the probability that the sum of the three outcomes is at least 6. It is easier to compute the probability of the complementary event that the sum is 5 or less, and then subtract from 1.

Given Data / Assumptions:

• We roll three fair six sided dice.
• Each die shows a number from 1 to 6.
• Total number of ordered outcomes = 6^3 = 216.
• We are interested in P(sum >= 6).

Concept / Approach:
Instead of counting all sums greater than or equal to 6 directly, we count the number of outcomes where the total is at most 5 and then use the complement. For three dice, only a few small combinations give sums 3, 4 or 5, so direct enumeration is manageable. Then P(sum at least 6) = 1 - P(sum at most 5).

Step-by-Step Solution:
Total outcomes = 6 * 6 * 6 = 216.Minimum sum is 1 + 1 + 1 = 3. So sums 3, 4 and 5 are the only ones less than 6.Count outcomes with sum = 3: only (1, 1, 1), so 1 outcome.Count outcomes with sum = 4: permutations of (1, 1, 2).Number of permutations of (1, 1, 2) = 3 (positions for the 2).Count outcomes with sum = 5: possibilities are (1, 1, 3) and (1, 2, 2).Permutations of (1, 1, 3) = 3; permutations of (1, 2, 2) = 3.So total outcomes with sum = 5 are 3 + 3 = 6.Total outcomes with sum at most 5 = 1 (sum 3) + 3 (sum 4) + 6 (sum 5) = 10.Therefore P(sum <= 5) = 10 / 216.Required probability P(sum >= 6) = 1 - 10 / 216 = 206 / 216 = 103 / 108 after simplification.

Verification / Alternative check:
We can verify the total small sum count by systematically listing partitions of 3, 4 and 5 into three positive integers between 1 and 6 and counting permutations. The result 10 is standard for three dice. Since the total number of outcomes is 216, the complement calculation is reliable. Additionally, note that 103/108 is close to 1, which makes sense because only a few outcomes have very small sums less than 6.

Why Other Options Are Wrong:
The fraction 103/216 would be the probability of the complementary event if we mistakenly considered 103 outcomes instead of 10, which is not correct. The fraction 103/36 is greater than 2 and cannot represent a probability. The ratio 36/103 is less than 1/2 and far too small, as only 10 out of 216 outcomes are excluded. Only 103/108 fits the correct complement calculation.

Common Pitfalls:
Students sometimes miscount the combinations for sums 4 and 5 or forget that (1, 1, 2) has three permutations, not one. Others attempt to compute the direct distribution of all sums from 3 to 18, which is unnecessary and error prone. Using the complement for small sums is much more efficient and reduces mistakes.

Final Answer:
The probability that the total is at least 6 is 103/108.