If x=3ⁿ, where n is a positive integral value, then what is the probability that x will have 3 at unit's place?

$\frac{1}{4}$

We have,

$3^1 = 3, 3^2 =9, 3^3 = 27,3^4 = 81, 3^5= 243, 3^6= 729$ etc.

So, a natural number of the form 3ⁿ may have either 3 or 9 or 7 or 1 at unit's place.

∴ Total number of elementary events = 4

Clearly, only one number has 3 at unit's place..

∴ Favourable number of ways = 1

Hence, required probability = $\frac{1}{4}$