An unbiased die with faces marked 1, 2, 3, 4, 5 and 6 is rolled four times. Out of four face values obtained the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5, is

$\frac{16}{81}$

P(minimum face value is not less than 2 and maximum face value is not greater than 5)

= P(2 or 3 or 4 or 5) $=\frac{4}{6}=\frac{2}{3}.$

Hence required probability $= {^4C}_4 \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^0 = \frac{16}{81}$.