Three identical dice are thrown together. The probability that distinct numbers appear on them, is

$\frac{4}{9}$ $\frac{5}{9}$ $\frac{5}{36}$ $\frac{1}{9}$

$\frac{5}{9}$

The total number of elementary events associated to the given random experiment is $6^3$. Three dice will show distinct numbers in ${^6C}_3 × 3!$ ways. ∴ Required probability =$\frac{^6C_3 × 3!}{6^3}=\frac{5}{9}$