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Consider a hash table with 100 slots. Collisions are resolved using chaining. Assuming simple uniform hashing, what is the probability that the first 3 slots are unfilled after the first 3 insertions?
  • a)
    (97 × 97 × 97) / 1003
  • b)
    (99 × 98 × 97) / 1003
  • c)
    (97 × 96 × 95) / 1003
  • d)
    (97 × 96 × 95) / (3! × 1003)
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Consider a hash table with 100 slots. Collisions are resolved using ch...
Chaining:
  • Each slot may contain a chain of elements but in linear probing, we can insert only one element in every slot.
  • In order to insert element 1, we have 4 to 100 = 97 slots out of 100 slots
  • In order to insert second element, we have 4 to 100 = 97 slots out of 100 slots
  • In order to insert third element, we have 4 to 100 = 97 slots out of 100 slots
  • Required probability 
Hence option a is the correct option.
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Community Answer
Consider a hash table with 100 slots. Collisions are resolved using ch...
The probability that the first slot is unfilled after the first insertion is 1, since all slots are initially unfilled.

After the first insertion, there is a probability of (99/100) that the second slot is unfilled, since there are 99 remaining unfilled slots out of the total 100 slots.

Similarly, after the second insertion, there is a probability of (98/100) that the third slot is unfilled.

Therefore, the probability that the first 3 slots are unfilled after the first 3 insertions is:
(1) * (99/100) * (98/100) = 0.9702

So the answer is approximately 0.9702, which is option a) 97%.
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Consider a hash table with 100 slots. Collisions are resolved using chaining. Assuming simple uniform hashing, what is the probability that the first 3 slots are unfilled after the first 3 insertions?a)(97 × 97 × 97) / 1003b)(99 × 98 × 97) / 1003c)(97 × 96 × 95) / 1003d)(97 × 96 × 95) / (3! × 1003)Correct answer is option 'A'. Can you explain this answer?
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