If the last digit must be a 3 and no digit can repeat, then there are 3 options (1, 2, and 4) for the first digit, 2 options for the second digit, 1 option for the third digit, and 1 option (a 3) for the last digit. The total number of outcomes that end with the number three is then \(\displaystyle{3}\times{2}\times{1}\times{1}={6}.\)

The total number of possible outcomes is \(\displaystyle{4}\times{3}\times{2}\times{1}={24}\) since there are 4 options for the first digit, 3 for the second, 2 for the third, and 1 for the fourth.

The probability an ID number will end with the number three is then \(\frac{6}{24}=\frac{1}{4}\)