Let `x = 33^n` . The index n is given a positive integral value at random. The probability that the value of x will have 3 in the units place is

Options

A. `(1)/(3)`
B. `(1)/(4)`
C. `(1)/(5)`
D. `(1)/(2)`

Answer

B

Explanation

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.
`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=`(1)/(4)`