1. **State the problem:** We want to find the probability that Xavier does not throw a 1 on his first throw but throws a 1 on his second throw with a fair six-sided die. 2. **Formula used:** This is a geometric probability problem where the probability of first success on the $k$-th trial is given by: $$P(X=k) = (1-p)^{k-1} p$$ where $p$ is the probability of success on each trial. 3. **Identify $p$:** Since the die is fair, the probability of throwing a 1 (success) on any throw is: $$p = \frac{1}{6}$$ 4. **Calculate the probability for $k=2$:** $$P(X=2) = (1-p)^{2-1} p = (1-\frac{1}{6})^{1} \times \frac{1}{6} = \frac{5}{6} \times \frac{1}{6}$$ 5. **Simplify the expression:** $$P(X=2) = \frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$$ 6. **Interpretation:** The probability that Xavier does not throw a 1 on the first throw but throws a 1 on the second throw is $\frac{5}{36}$. **Final answer:** $$\boxed{\frac{5}{36}}$$