Say you choose one of them. There remain 7 possible digits for the last 3 numbers.
Let's take care of the 3rd. There are 7 possible.
Let's take care of the 2nd. There are 6 possible.
Let's take care of the 1st. There are 5 possible.
The numbers of possibilities is hence
The overall number of possibilities is (no repetition, so each time you pick a number, you can't pick it again).
So the probability is
Actually, it could have been more straightforward, because there are as many even numbers as odd numbers that are 4-digit without repetition !
I'll try the other ones later