If four fair six-sided dice are rolled, what is the probability that the lowest number appearing on any die is exactly 3 ?
The probability that all the die rolls are at least 3 is $\frac{4}{6}$. The probability they are all at least 4 is $\frac{3}{6}$. The probability of being in the former category but not the latter is thus $\frac{4}{6}^{4}-\frac{3}{6}^{4}=\frac{256-81}{1296}=\frac{175}{1296}$. $\fbox{175 / 1296}$.