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

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 $3^{n}$ 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= $= \frac{1}{4}$

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