The probability of obtaining a 1 or a 6 on a single throw of a fair die is \(\frac{2}{6} = \frac{1}{3}\).

The probability of not obtaining a 1 or a 6 on a single throw is \(\frac{4}{6} = \frac{2}{3}\).

We need to find the probability that it takes at least 3 throws but no more than 5 throws to obtain a 1 or a 6.

This can be calculated as:

\(\left( \frac{1}{3} \right) \left( \frac{2}{3} \right)^2 + \left( \frac{1}{3} \right) \left( \frac{2}{3} \right)^3 + \left( \frac{1}{3} \right) \left( \frac{2}{3} \right)^4\)

Calculating each term:

\(\left( \frac{1}{3} \right) \left( \frac{2}{3} \right)^2 = \frac{4}{27}\)

\(\left( \frac{1}{3} \right) \left( \frac{2}{3} \right)^3 = \frac{8}{81}\)

\(\left( \frac{1}{3} \right) \left( \frac{2}{3} \right)^4 = \frac{16}{243}\)

Adding these probabilities:

\(\frac{4}{27} + \frac{8}{81} + \frac{16}{243} = \frac{76}{243}\)

Thus, the probability is \(\frac{76}{243}\) or approximately 0.313.