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 is getting a 1 or a 6. The probability of success (getting 1 or 6) on any throw is $p = \frac{2}{6} = \frac{1}{3}$. 3. **Probability of failure:** The probability of not getting 1 or 6 on a throw is $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$. 4. **Distribution:** The number of throws until the first success follows a geometric distribution with parameter $p$. 5. **We want:** $P(3 \leq X \leq 5)$ where $X$ is the trial on which the first success occurs. 6. **Formula for geometric distribution:** $$ P(X = k) = q^{k-1} p $$ 7. **Calculate:** $$ P(3 \leq X \leq 5) = P(X=3) + P(X=4) + P(X=5) = q^{2}p + q^{3}p + q^{4}p $$ 8. **Substitute values:** $$ = \left(\frac{2}{3}\right)^2 \cdot \frac{1}{3} + \left(\frac{2}{3}\right)^3 \cdot \frac{1}{3} + \left(\frac{2}{3}\right)^4 \cdot \frac{1}{3} $$ 9. **Simplify:** $$ = \frac{1}{3} \left( \left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^3 + \left(\frac{2}{3}\right)^4 \right) = \frac{1}{3} \left( \frac{4}{9} + \frac{8}{27} + \frac{16}{81} \right) $$ 10. **Find common denominator 81:** $$ = \frac{1}{3} \left( \frac{36}{81} + \frac{24}{81} + \frac{16}{81} \right) = \frac{1}{3} \cdot \frac{76}{81} = \frac{76}{243} $$ **Final answer:** $$ P(3 \leq X \leq 5) = \frac{76}{243} \approx 0.3127 $$