A fair die is rolled 5 times. What is the probability of having no 1 and no 5 among the rolls? Round your answer to three decimal places.

To find the probability of rolling no 1s and no 5s when rolling a fair die 5 times, we first determine the possible outcomes for a single roll. Since we're avoiding 1s and 5s, we only have the numbers 2, 3, 4, and 6 available — that's 4 possible outcomes. The total number of outcomes for one roll of a die is 6. Therefore, the probability of rolling either a 1 or a 5 is: \[ P(\text{1 or 5}) = \frac{2}{6} = \frac{1}{3} \] Thus, the probability of not rolling a 1 or a 5 (rolling a 2, 3, 4, or 6) in a single roll is: \[ P(\text{not 1 and not 5}) = 1 - P(\text{1 or 5}) = 1 - \frac{1}{3} = \frac{2}{3} \] Next, since the die is rolled 5 times, we raise the probability of a single successful outcome to the power of the number of rolls: \[ P(\text{no 1s and no 5s in 5 rolls}) = \left(\frac{2}{3}\right)^5 \] Calculating this gives: \[ P = \left(\frac{2}{3}\right)^5 = \frac{32}{243} \] To find this probability as a decimal, we calculate: \[ \frac{32}{243} \approx 0.131687 \] Rounding this to three decimal places, we get: **0.132**