Four whole numbers are randomly selected and are multiplied. Let us define
E₁ : Last digit in the product is either 1 or 3 or 7 or 9
E_2: Last digit in the product is either 2 or 4 or 6 or 8 then

The last digit in the product ‘a’ will be 1 or 3 or 7 or 9 only when the last digit in each of the ‘n’ numbers ${\mathrm{x}}_1,{\mathrm{x}}_2,\dots \dots {\mathrm{x}}_{\mathrm{n}}$ is 1 or 3 or 7 or 9.
$\mathrm{n}\left({\mathrm{E}}_1\right)=4^4\mathrm{P}\left({\mathrm{E}}_1\right)=\frac{4^4}{{10}^4}={\left(\frac{2}{5}\right)}^4$
E₃ = the event that the last digit in the number a
is 1,2,3,4,5,6,7,8,9  n(E₃ ) 8⁴
$\begin{array}{l}\mathrm{n}\left({\mathrm{E}}_2\right)=\mathrm{n}\left({\mathrm{E}}_3\right)-\mathrm{n}\left({\mathrm{E}}_1\right)=8^4-4^4\\ \mathrm{P}\left({\mathrm{E}}_2\right)=\frac{8^4-4^4}{{10}^4}=\frac{4^4-2^4}{5^4}\end{array}$