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. **Identify the distribution:** This is a binomial probability problem where each roll is a trial with two outcomes: rolling a 1 (success) or not rolling a 1 (failure). 3. **Parameters:** Number of trials $n=6$, number of successes $k=1$, probability of success on each trial $p=\frac{1}{6}$, probability of failure $q=1-p=\frac{5}{6}$. 4. **Binomial probability formula:** $$P(X=k) = \binom{n}{k} p^k q^{n-k}$$ 5. **Calculate the binomial coefficient:** $$\binom{6}{1} = 6$$ 6. **Calculate the probability:** $$P(X=1) = 6 \times \left(\frac{1}{6}\right)^1 \times \left(\frac{5}{6}\right)^5$$ 7. **Simplify:** $$P(X=1) = 6 \times \frac{1}{6} \times \left(\frac{5}{6}\right)^5 = 1 \times \left(\frac{5}{6}\right)^5 = \left(\frac{5}{6}\right)^5$$ 8. **Final answer:** $$P(X=1) = \left(\frac{5}{6}\right)^5 \approx 0.4019$$ So, the probability that the number 1 is rolled exactly once in six rolls is approximately 0.4019.