As chaining is used to resolve collision
A. The first element can go anywhere except the first block so P(getting into other boxes except first)=n-1/n
Here block A continues to be empty so probability = n-1/n * n-1/n * ......k times = (n-1/n)^k
B. No collision in k insertions so 1st element can go anywhere similarly 2nd can go to any block except the 1st block
so, probability = 1 * n-1/n * n-2/n * ......* n-k+1/n
C. If 1st collision occurs at k th insertion then probability = 1 * n-1/n * n-2/n * ...... * n-k+2/n * k-1/n