A fair six-sided die is rolled three times. What is the probability that the face with number 6 does not appear on all three rolls (that is, it is not the case that all three rolls show 6)?

Introduction / Context:
This problem involves repeated rolling of a fair six-sided die and asks for the probability that the outcome is not six on all three rolls simultaneously. In other words, we want the probability that it is not true that each of the three rolls shows a 6. This is a classic use of the complement rule in probability.

Given Data / Assumptions:

• A fair six-sided die is rolled three times.
• Each roll is independent.
• Each face (1 through 6) has probability 1/6 on any single roll.
• We are interested in the event that not all three rolls show 6.

Concept / Approach:
It is easier to compute the probability of the complementary event where all three rolls show 6, and then subtract this from 1. The event "all three rolls are 6" is straightforward to evaluate because the rolls are independent. Once that is known, the required probability is 1 minus this value.

Step-by-Step Solution:
Probability that one roll shows 6 is 1/6. Because the rolls are independent, probability that all three rolls show 6 is (1/6) * (1/6) * (1/6) = 1 / 6^3 = 1 / 216. Let A be the event that all three rolls are 6. We need the probability of not A, that is, the event that not all three rolls are 6. P(not A) = 1 - P(A). So P(not all three are 6) = 1 - 1/216 = (216/216) - (1/216) = 215/216.

Verification / Alternative check:
We can check that this is very close to 1, which is reasonable, because the event that all three rolls are 6 is extremely rare. The chance 1/216 is small, so its complement 215/216 is large. Also, the result is consistent with the idea that only one outcome out of the 216 total ordered triplets corresponds to (6,6,6), while all other 215 outcomes belong to the desired event.

Why Other Options Are Wrong:
1/216: This is the probability that all three rolls are 6, which is the complementary event, not the required one.
215/256: This denominator 256 is not appropriate for three rolls of a six-sided die, because the correct total number of ordered outcomes is 6^3 = 216.
1/52: This number is unrelated to this experiment and likely comes from thinking about cards, not dice.
None of these: This is incorrect because 215/216 is listed among the options and is correct.

Common Pitfalls:
A usual misunderstanding is to interpret the question as "no 6 appears at all", which would give probability (5/6)^3 instead of the complement of all three being 6. Reading the wording carefully is essential. Another mistake is miscalculating 6^3 or forgetting to use the complement rule when it offers a simpler solution.

Final Answer:
Thus, the probability that it is not the case that all three rolls show 6 is 215/216.