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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?
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Verified Answer
Consider a hash table with 100 slots. Collisions are resolved using ch...
Simple Uniform hashing function is a hypothetical hashing function that evenly distributes items into the slots of a hash table. Moreover, each item to be hashed has an equal probability of being placed into a slot, regardless of the other elements already placed.
Probability that the first 3 slots are unfilled after the first 3 insertions =
(probability that first item doesn't go in any of the first 3 slots)*
(probability that second item doesn't go in any of the first 3 slots)*
(probability that third item doesn't go in any of the first 3 slots)
(probability that first item doesn't go in any of the first 3 slots)*
(probability that second item doesn't go in any of the first 3 slots)*
(probability that third item doesn't go in any of the first 3 slots)
= (97/100) * (97/100) * (97/100)
Most Upvoted Answer
Consider a hash table with 100 slots. Collisions are resolved using ch...
Given information:
- Hash table with 100 slots
- Collisions resolved using chaining
- Simple uniform hashing
To find: Probability that the first 3 slots are unfilled after the first 3 insertions
Solution:
1. Total number of possible outcomes:
- Each insertion can be made into any of the 100 slots independently.
- Therefore, the total number of possible outcomes for the first insertion is 100.
- Similarly, for the second and third insertions, the number of possible outcomes is also 100 each.
- Since these events are independent, the total number of possible outcomes for all 3 insertions is 100 * 100 * 100 = 100^3.
2. Number of favorable outcomes:
- To have the first 3 slots unfilled, each insertion must land in one of the remaining unfilled slots.
- After the first insertion, there are 100 slots available, so the probability of landing in an unfilled slot is 100/100 = 1.
- After the second insertion, there are 99 slots available (1 slot was filled in the previous step), so the probability of landing in an unfilled slot is 99/99 = 1.
- Similarly, after the third insertion, there are 98 slots available, so the probability of landing in an unfilled slot is 98/98 = 1.
- Therefore, the number of favorable outcomes is 1 * 1 * 1 = 1.
3. Calculating the probability:
- Probability is defined as the ratio of favorable outcomes to total outcomes.
- In this case, the probability is 1/100^3 = 1/100^3 = 1/1000000.
- Simplifying the fraction, we get 1/1000000 = 1/1000 * 1/1000 = 1/1000^2 = 1/1000 * 1/1000.
- Therefore, the probability is (1/1000) * (1/1000) * (1/1000).
Final Answer:
- The probability that the first 3 slots are unfilled after the first 3 insertions is (1/1000) * (1/1000) * (1/1000).
- Simplifying, this can be written as (1/1000)^3 = 1/1000000000 = 1/1003.
- Therefore, the correct answer is option A) (97 97 97)/1003.
- Hash table with 100 slots
- Collisions resolved using chaining
- Simple uniform hashing
To find: Probability that the first 3 slots are unfilled after the first 3 insertions
Solution:
1. Total number of possible outcomes:
- Each insertion can be made into any of the 100 slots independently.
- Therefore, the total number of possible outcomes for the first insertion is 100.
- Similarly, for the second and third insertions, the number of possible outcomes is also 100 each.
- Since these events are independent, the total number of possible outcomes for all 3 insertions is 100 * 100 * 100 = 100^3.
2. Number of favorable outcomes:
- To have the first 3 slots unfilled, each insertion must land in one of the remaining unfilled slots.
- After the first insertion, there are 100 slots available, so the probability of landing in an unfilled slot is 100/100 = 1.
- After the second insertion, there are 99 slots available (1 slot was filled in the previous step), so the probability of landing in an unfilled slot is 99/99 = 1.
- Similarly, after the third insertion, there are 98 slots available, so the probability of landing in an unfilled slot is 98/98 = 1.
- Therefore, the number of favorable outcomes is 1 * 1 * 1 = 1.
3. Calculating the probability:
- Probability is defined as the ratio of favorable outcomes to total outcomes.
- In this case, the probability is 1/100^3 = 1/100^3 = 1/1000000.
- Simplifying the fraction, we get 1/1000000 = 1/1000 * 1/1000 = 1/1000^2 = 1/1000 * 1/1000.
- Therefore, the probability is (1/1000) * (1/1000) * (1/1000).
Final Answer:
- The probability that the first 3 slots are unfilled after the first 3 insertions is (1/1000) * (1/1000) * (1/1000).
- Simplifying, this can be written as (1/1000)^3 = 1/1000000000 = 1/1003.
- Therefore, the correct answer is option A) (97 97 97)/1003.
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Consider a hash table with 100 slots. Collisions are resolved using ch...
B
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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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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? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about 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? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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?.
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? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about 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? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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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