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Given a hash table with n keys and m slots with simple uniform hashing. If collisions are resolved by chaining then what is the probability that the first slot ends up empty ?
  • a)
    (1 / m)n
  • b)
    [1 – (1/m)]n
  • c)
    (1/n)m
  • d)
    [1 – (1/n)]m
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Given a hash table with n keys and m slots with simple uniform hashing...
​Probability of one particular slot = 1/m ( because total m slots)
Probability that a value should not go in one particular slot = 1 – (1/m)
For n values (keys) probability = [1 – (1/m)]n
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Given a hash table with n keys and m slots with simple uniform hashing. If collisions are resolved by chaining then what is the probability that the first slot ends up empty ?a)(1 / m)nb)[1 – (1/m)]nc)(1/n)md)[1 – (1/n)]mCorrect answer is option 'B'. Can you explain this answer?
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