To make the table 6% full, we need to insert at least ( 75∙6 100) = 4.5 round up to 5 values.
Probability of collision during first insertion is 1/ 75
Probability of collision during third insertion is 2 /75
Probability of collision during fourth insertion is 3/ 75
Probability of collision during fifth insertion is 4 /75
Probability of collision during sixth insertion is 5 /75
So, total probability of collision to make the table 6% full is (1 + 2 + 3 + 4 + 5) /75 = 0.2
So option B is correct