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2 Answers

5 votes
5 votes

A hash table has space for 75 records, then the probability of collision before the table is 6% full.

So collision will happen before $75\times 0.6 % \approx 5$  ie. collision happen before 5 insertion.

Probability that collision happen before 5 = 1 - no collision Till  5 

= 1 - $\left ( \frac{75\times 74\times 73\times 72\times 71}{75^{5}} \right )$

= 1 - 0.87

= 0.127

–1 votes
–1 votes
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

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