@Manoj_Kumar you can solve this way using your approach :
For that 3 cases are possible
Case 1- Collission on First insertion and Not Collision on Second Insertion
P(case-1) = (30/100)* (69/100) = 0.207
Case 2- Not Collission on First insertion and Collision on Second Insertion
P(case-2) = (70/100)* (31/100) =0.217
Case 3- Collission on both First insertion and Second Insertion
P(case-3) = (30/100)*(31/100) = 0.093
Finally,
Required Probability = P(Case-1) or P(Case-2) or P(Case-3)
=0.207 + 0.217 + 0.093 = 0.517