Probability, Statistics, Distributed Systems, Software Engineering, Interview Questions.
put along with the 1st key since it will lead to collision. Probability of the 2nd key having no collision after 1st key inserted = N-1/N Similarly, after the 2nd key is put, the 3rd key has N-2 choices left (excluding the slots for the 1st and 2nd key). Probability of the 3rd key having no