Answer : Option D

Explanation :

The numbers can be written in 5! different ways then n(S) = 5! = 120
Let E be the event that getting 4 digit number with 1 in the unit place and 2 in the tens place when the numbers 1,2,3,4 and 5 are arranged at random.
 Since each desired number is ending with 1,there is only 1 way.
The Second place(tens) can now be filled with 2.
Here also only 1 way to fill.
The next place(hundreds) can now be filled by any of the remaining 3 numbers.
So, there are 3 ways to filling that place.
Then, the first place can now be filled by any of the remaining 2 numbers.
So, there are 2 ways to fill.
Therefore n(E) = 1 x 1 x 3 x 2 = 6.
Now, p(E) = n(E) / n(S) = 6 / 120 = 1/20.
Hence the required probability is 1/20.