Data structures (C++)
please explain and solve it step by step
Suppose that you are using the following hash function
to keep records of students in hash map.
H(K) = month of K
where K is the birth date of the student.
What is the probability that you will have at least one
collision after inserting 50 students in the databese?
Explain and show your calculations clearly.

Question

Data structures (C++)
please explain and solve it step by step
Suppose that you are using the following hash function
to keep records of students in hash map.
H(K) = month of K
where K is the birth date of the student.
What is the probability that you will have at least one
collision after inserting 50 students in the databese?
Explain and show your calculations clearly.

Accepted Answer

⚹ To solve the problem of calculating the probability of at least one collision when inserting 50 students into a database using a hash function that maps the birth date of a student to the month of birth, we can use the concept of the birthday problem in probability theory. ⚹ ⚹ The birthday problem traditionally asks for the probability that, in a set of n randomly chosen people, at least two will have the same birthday. In this case, we are looking at months instead of days, so there are 12 possible outcomes for each student's hash value. ⚹ ⚹ The probability of no collision (i.e., all hash values are unique) when inserting the second student is $\frac{11}{12}$, since there are 11 months left that haven't been chosen by the first student. ⚹ ⚹ Similarly, the probability of no collision when inserting the third student is $\frac{10}{12}$, as there are now 10 months left. ⚹ ⚹ Continuing this pattern, the probability of no collision for the 50th student is $\frac{12 - 49}{12}$, but since this is not possible (as the probability would be negative), we can see that a collision is certain by the time we try to insert the 13th student. ⚹ ⚹ However, we are interested in the probability of at least one collision after inserting 50 students, so we need to calculate the probability of no collision for the first 12 students and subtract this from 1. ⚹ ⚹ The probability of no collision for the first 12 students can be calculated as follows: ⚹ $$ P( ext{no collision}) = \frac{12}{12} imes \frac{11}{12} imes \frac{10}{12} imes \ldots imes \frac{1}{12} $$ ⚹ Simplifying the above expression, we get: ⚹ $$ P( ext{no collision}) = \frac{12!}{12^{12}} $$ ⚹ Therefore, the probability of at least one collision, $P( ext{at least one collision})$, is: ⚹ $$ P( ext{at least one collision}) = 1 - P( ext{no collision}) $$ $$ P( ext{at least one collision}) = 1 - \frac{12!}{12^{12}} $$ ⚹ Calculating the above expression gives us the probability of having at least one collision after inserting 50 students into the database. ⚹ ⚹ Since the number of students (50) exceeds the number of possible hash values (12), a collision is guaranteed, and thus the probability of at least one collision is 1. ⚹

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