If n positive integers are taken at random and multiplied together, the probability that the last digit of the product is 2, 4, 6 or 8, is

The last digit of the product will be 1,2,3,4,5,6,7,8 or 9 if and only if each of the n positive integers ends in any of these digits. Now the probability of an integer ending in 1,2,3,4,5,6,7,8 or 9 is $\frac{8}{10}$ There fore the probability that the last digit of the product of $n$ integers in $1,2,3,4,5,6,7,8$  or $9$is  ${\left(\frac{4}{5}\right)}^n$ The probability for an interger to end in 1,3,7 or 9 is  $\frac{4}{10}=\frac{2}{5}$
There fore the probability for the product of $n$ positive integers to end in 1,3,7 or 9 is  ${\left(\frac{2}{5}\right)}^n$
Hence the required probability 

$={\left(\frac{4}{5}\right)}^n-{\left(\frac{2}{5}\right)}^n=\frac{4^n-2^n}{5^n}$