The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened. In this case, event A is selecting one defective item and event B is selecting another defective item.
The probability of selecting one defective item is the number of defective items divided by the total number of items, $\frac{20}{100}$.
The probability of selecting another defective item after one defective item has already been selected is the number of defective items remaining divided by the total number of items remaining, $\frac{19}{99}$.
Therefore, the probability of selecting two defective items at random without replacement is $\frac{20}{100} \cdot \frac{19}{99} = \boxed{\frac{19}{495}}$.
Option A is incorrect because it is the probability of selecting one defective item. Option B is incorrect because it is the probability of selecting two defective items with replacement. Option C is incorrect because it is the probability of selecting one defective item and one non-defective item.