1. **Problem statement:** We roll a fair six-sided die repeatedly until we get a 1 or a 6. We want to find the probability that it takes at least 3 throws but no more than 5 throws to get a 1 or a 6. 2. **Understanding the problem:** The die has 6 faces, and the event of interest (getting a 1 or 6) has probability $p = \frac{2}{6} = \frac{1}{3}$ on each throw. 3. **Complement event:** The probability of not getting a 1 or 6 on a single throw is $q = 1 - p = \frac{2}{3}$. 4. **Distribution:** The number of throws until the first success (getting 1 or 6) follows a geometric distribution with parameter $p = \frac{1}{3}$. 5. **Probability that it takes exactly $k$ throws:** $$P(X = k) = q^{k-1} p = \left(\frac{2}{3}\right)^{k-1} \cdot \frac{1}{3}$$ 6. **Probability that it takes at least 3 but no more than 5 throws:** $$P(3 \leq X \leq 5) = P(X=3) + P(X=4) + P(X=5)$$ 7. **Calculate each term:** $$P(X=3) = \left(\frac{2}{3}\right)^2 \cdot \frac{1}{3} = \frac{4}{9} \cdot \frac{1}{3} = \frac{4}{27}$$ $$P(X=4) = \left(\frac{2}{3}\right)^3 \cdot \frac{1}{3} = \frac{8}{27} \cdot \frac{1}{3} = \frac{8}{81}$$ $$P(X=5) = \left(\frac{2}{3}\right)^4 \cdot \frac{1}{3} = \frac{16}{81} \cdot \frac{1}{3} = \frac{16}{243}$$ 8. **Sum the probabilities:** $$P(3 \leq X \leq 5) = \frac{4}{27} + \frac{8}{81} + \frac{16}{243}$$ 9. **Find common denominator 243:** $$\frac{4}{27} = \frac{36}{243}, \quad \frac{8}{81} = \frac{24}{243}, \quad \frac{16}{243} = \frac{16}{243}$$ 10. **Add:** $$\frac{36}{243} + \frac{24}{243} + \frac{16}{243} = \frac{76}{243}$$ **Final answer:** $$\boxed{\frac{76}{243}}$$ This is the probability that it takes at least 3 but no more than 5 throws to get a 1 or a 6.