Three fair dice are thrown simultaneously. What is the probability that at least one of the dice shows a six?

Introduction / Context:
This question uses the complement rule in probability for repeated independent trials. We throw three fair dice at the same time and want the probability that at least one die shows the number six. Directly counting arrangements with one or more sixes can be messy. Instead, it is much easier to calculate the probability that no die shows a six and subtract this from 1 to obtain the probability of at least one six.

Given Data / Assumptions:

Three fair six faced dice are thrown simultaneously.

Each die has faces numbered 1 to 6.

The dice are independent, and each face on each die has probability 1/6.

We are interested in the event that at least one die shows the face 6.

Concept / Approach:
The complement of the event "at least one six" is the event "no six appears". If we can find the probability that none of the three dice shows a six, we can subtract this probability from 1. This method significantly reduces counting errors and is standard for "at least one" type problems in probability.

Step-by-Step Solution:
Step 1: For a single fair die, probability that it does not show a six is 5/6.Step 2: Because the dice are independent, the probability that all three dice show numbers that are not six is (5/6)^3.Step 3: Compute (5/6)^3 = (5 * 5 * 5) / (6 * 6 * 6) = 125 / 216.Step 4: This is the probability of the complement event (no six at all).Step 5: Probability of at least one six = 1 - probability of no six.Step 6: Therefore, P(at least one six) = 1 - 125 / 216 = (216 - 125) / 216 = 91 / 216.

Verification / Alternative check:
We can also check by computing probability of exactly one six, exactly two sixes and exactly three sixes and then summing them. For example, P(exactly one six) = 3 * (1/6) * (5/6)^2, P(exactly two sixes) = 3 * (1/6)^2 * (5/6), and P(exactly three sixes) = (1/6)^3. Summing these three values also gives 91 / 216 after simplification. This verifies that the complement based calculation is correct.

Why Other Options Are Wrong:
The value 1/36 is much too small and does not reflect the many ways at least one six can occur on three dice. The fraction 13/216 is too small and could correspond to a poorly calculated partial event. The value 125/216 is actually the probability of the complement event (no six on any die), not the desired event. The fraction 5/36 is also incorrect and does not correspond to any natural counting pattern for this experiment.

Common Pitfalls:
A common mistake is to treat "at least one six" as equivalent to "exactly one six" and only count those cases. Another frequent error is misusing addition or not recognising independence, leading to wrong exponents in the calculation. Using the complement rule and clearly identifying the no six event as (5/6)^3 is the safest route.

Final Answer:
The probability that at least one of the three dice shows a six is 91/216.