1. **State the problem:** We want to find the probability that when a six-sided die is rolled six times, the number 1 appears exactly once. 2. **Formula used:** This is a binomial probability problem. The probability of exactly $k$ successes in $n$ independent trials is given by: $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $p$ is the probability of success on a single trial. 3. **Identify parameters:** - Number of trials $n = 6$ - Number of successes $k = 1$ - Probability of rolling a 1 on a single roll $p = \frac{1}{6}$ 4. **Calculate binomial coefficient:** $$\binom{6}{1} = 6$$ 5. **Calculate probability:** $$P(X=1) = 6 \times \left(\frac{1}{6}\right)^1 \times \left(1 - \frac{1}{6}\right)^{5} = 6 \times \frac{1}{6} \times \left(\frac{5}{6}\right)^5$$ 6. **Simplify:** $$6 \times \frac{1}{6} = 1$$ So, $$P(X=1) = \left(\frac{5}{6}\right)^5$$ 7. **Final answer:** $$P(X=1) = \left(\frac{5}{6}\right)^5 \approx 0.4019$$ This means there is about a 40.19% chance that the number 1 is rolled exactly once in six rolls of a six-sided die.