A number x is chosen from the set A = {$33^n$: n = N}. The probability that x has 3 in units place, is

$\frac{1}{4}$

We observe that the numbers in the set A have either 3 or 9 or 7 or 1 at units place. Thus, the numbers in set A can be grouped into four groups viz. numbers having 1 at units place, numbers having 3 at units place, numbers having 7 at units place and numbers having 9 at units place.

∴ Required probability $=\frac{1}{4}$