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A hash function randomly distributes records one by one in a space that can hold x number of records. The probability that the mth record is the first record to result in collision is
- a)(x - 1) (x - 2)...(X - (m - 2)) (m - 1) / Xm-1
- b)(x - 1) (x - 2)...(X - (m - 1)) (m - 1) / Xm-1
- c)(x - 1) (x - 2)...(X - (m - 2)) (m - 1) / Xm
- d)(x - 1) (x - 2)...(X - (m - 2)) (m - 1) / Xm+1
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A hash function randomly distributes records one by one in a space tha...
Probability for the first record not colliding is x/x.
Probability for the second record not colliding is x - 1/x. (This is because one place is already occupied. So favorable number of cases is x-1).
Probability for the third record not colliding is x - 2/x.
Probability for the (m - 1)th record not colliding is x - (m - 2)/x.
Now the next (mth) record is resulting in a collision. Out of the x places, it should hash to one of the (m - 1) places already, filled. So probability is (m - 1)/x.
The required probability is (x/x) (x - 1/x) (x - 2/x) ...(x - (m - 2)/x) (m - 1/x)
Probability for the second record not colliding is x - 1/x. (This is because one place is already occupied. So favorable number of cases is x-1).
Probability for the third record not colliding is x - 2/x.
Probability for the (m - 1)th record not colliding is x - (m - 2)/x.
Now the next (mth) record is resulting in a collision. Out of the x places, it should hash to one of the (m - 1) places already, filled. So probability is (m - 1)/x.
The required probability is (x/x) (x - 1/x) (x - 2/x) ...(x - (m - 2)/x) (m - 1/x)
Most Upvoted Answer
A hash function randomly distributes records one by one in a space tha...
The Answer Explained:
To understand why option 'A' is the correct answer, let's break down the probability calculation step by step.
Step 1: Probability of the first record not causing a collision
When the first record is inserted into the hash space, it does not collide with any existing records. The probability of this happening is 1 (since there are no other records to collide with).
Step 2: Probability of the second record not causing a collision
When the second record is inserted, there is already one record in the hash space. The probability of the second record not causing a collision is (x-1)/x. This is because there are x-1 empty slots available out of x total slots, and the second record should not occupy the same slot as the first record.
Step 3: Probability of the third record not causing a collision
When the third record is inserted, there are already two records in the hash space. The probability of the third record not causing a collision is (x-2)/x. This is because there are x-2 empty slots available out of x total slots, and the third record should not occupy the same slot as the first two records.
Step m: Probability of the mth record not causing a collision
Following the same logic, when the mth record is inserted, there are already (m-1) records in the hash space. The probability of the mth record not causing a collision is (x-(m-1))/x. This is because there are x-(m-1) empty slots available out of x total slots, and the mth record should not occupy the same slot as the previous (m-1) records.
Step (m+1): Probability of the (m+1)th record causing a collision
Now, when the (m+1)th record is inserted, there are already m records in the hash space. The probability of the (m+1)th record causing a collision is 1 - [(x-(m-1))/x]. This is because the (m+1)th record can only cause a collision if it occupies the same slot as one of the previous m records.
Overall probability:
To calculate the overall probability that the mth record is the first record to result in a collision, we multiply the probabilities from step 1 to step (m+1) together. This can be represented as:
1 * ((x-1)/x) * ((x-2)/x) * ... * (x-(m-2))/(x * (x-(m-1))/x)
Simplifying this expression, we get:
(x-1) * (x-2) * ... * (x-(m-2)) * (m-1) / xm-1
Therefore, the correct answer is option 'A': (x-1) * (x-2) * ... * (x-(m-2)) * (m-1) / xm-1.
To understand why option 'A' is the correct answer, let's break down the probability calculation step by step.
Step 1: Probability of the first record not causing a collision
When the first record is inserted into the hash space, it does not collide with any existing records. The probability of this happening is 1 (since there are no other records to collide with).
Step 2: Probability of the second record not causing a collision
When the second record is inserted, there is already one record in the hash space. The probability of the second record not causing a collision is (x-1)/x. This is because there are x-1 empty slots available out of x total slots, and the second record should not occupy the same slot as the first record.
Step 3: Probability of the third record not causing a collision
When the third record is inserted, there are already two records in the hash space. The probability of the third record not causing a collision is (x-2)/x. This is because there are x-2 empty slots available out of x total slots, and the third record should not occupy the same slot as the first two records.
Step m: Probability of the mth record not causing a collision
Following the same logic, when the mth record is inserted, there are already (m-1) records in the hash space. The probability of the mth record not causing a collision is (x-(m-1))/x. This is because there are x-(m-1) empty slots available out of x total slots, and the mth record should not occupy the same slot as the previous (m-1) records.
Step (m+1): Probability of the (m+1)th record causing a collision
Now, when the (m+1)th record is inserted, there are already m records in the hash space. The probability of the (m+1)th record causing a collision is 1 - [(x-(m-1))/x]. This is because the (m+1)th record can only cause a collision if it occupies the same slot as one of the previous m records.
Overall probability:
To calculate the overall probability that the mth record is the first record to result in a collision, we multiply the probabilities from step 1 to step (m+1) together. This can be represented as:
1 * ((x-1)/x) * ((x-2)/x) * ... * (x-(m-2))/(x * (x-(m-1))/x)
Simplifying this expression, we get:
(x-1) * (x-2) * ... * (x-(m-2)) * (m-1) / xm-1
Therefore, the correct answer is option 'A': (x-1) * (x-2) * ... * (x-(m-2)) * (m-1) / xm-1.
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A hash function randomly distributes records one by one in a space that can hold x number of records. The probability that the mth record is the first record to result in collision isa)(x- 1) (x - 2)...(X - (m - 2)) (m - 1) / Xm-1b)(x- 1) (x - 2)...(X - (m - 1)) (m - 1) / Xm-1c)(x- 1) (x - 2)...(X - (m - 2)) (m - 1) / Xmd)(x- 1) (x - 2)...(X - (m - 2)) (m - 1) / Xm+1Correct answer is option 'A'. Can you explain this answer?
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A hash function randomly distributes records one by one in a space that can hold x number of records. The probability that the mth record is the first record to result in collision isa)(x- 1) (x - 2)...(X - (m - 2)) (m - 1) / Xm-1b)(x- 1) (x - 2)...(X - (m - 1)) (m - 1) / Xm-1c)(x- 1) (x - 2)...(X - (m - 2)) (m - 1) / Xmd)(x- 1) (x - 2)...(X - (m - 2)) (m - 1) / Xm+1Correct 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 A hash function randomly distributes records one by one in a space that can hold x number of records. The probability that the mth record is the first record to result in collision isa)(x- 1) (x - 2)...(X - (m - 2)) (m - 1) / Xm-1b)(x- 1) (x - 2)...(X - (m - 1)) (m - 1) / Xm-1c)(x- 1) (x - 2)...(X - (m - 2)) (m - 1) / Xmd)(x- 1) (x - 2)...(X - (m - 2)) (m - 1) / Xm+1Correct 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 A hash function randomly distributes records one by one in a space that can hold x number of records. The probability that the mth record is the first record to result in collision isa)(x- 1) (x - 2)...(X - (m - 2)) (m - 1) / Xm-1b)(x- 1) (x - 2)...(X - (m - 1)) (m - 1) / Xm-1c)(x- 1) (x - 2)...(X - (m - 2)) (m - 1) / Xmd)(x- 1) (x - 2)...(X - (m - 2)) (m - 1) / Xm+1Correct answer is option 'A'. Can you explain this answer?.
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ample number of questions to practice A hash function randomly distributes records one by one in a space that can hold x number of records. The probability that the mth record is the first record to result in collision isa)(x- 1) (x - 2)...(X - (m - 2)) (m - 1) / Xm-1b)(x- 1) (x - 2)...(X - (m - 1)) (m - 1) / Xm-1c)(x- 1) (x - 2)...(X - (m - 2)) (m - 1) / Xmd)(x- 1) (x - 2)...(X - (m - 2)) (m - 1) / Xm+1Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Computer Science Engineering (CSE) tests.
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