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

# 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?

1. A

$\frac{1}{3}$

2. B

$\frac{1}{4}$

3. C

$\frac{1}{5}$

4. D

$\frac{1}{2}$

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### Solution:

We have

etc.

So, a natural number of the form ${3}^{n}$ may have either

3 or 9 or 7 or 1 at unit's place.

$\therefore$  Total number of elementary events= 4

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

$\therefore$ Favourable number of ways = 1

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

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