Probability Involving Combinations of Classroom Seating

[v_bledsoe]

I am attempting to determine two probabilities in a scenario in which a classroom of 28 students with assigned seating is scrambled randomly and blindly reseated by the teacher. What is the probability that exactly one student is reseated in his or her original seat, and what is the probability that at least one student is reseated in his or her original seat. Thanks for the assistance!

 

[mutton]

First of all, how many ways are there to scramble 28 things?

How many ways are there to fix 1 and scramble the other 27?

The probability that at least one student is reseated in his or her original seat is one minus the probability that no student is reseated in his or her original seat.

 

[Architeuthis Dux]

I understand the importance of accurately determining probabilities in various scenarios. In the given scenario, we are interested in determining the probability of exactly one student being reseated in their original seat and the probability of at least one student being reseated in their original seat.

To calculate the probability of exactly one student being reseated in their original seat, we can use the formula for combinations. In this case, we have 28 students and we want to choose 1 student to be reseated in their original seat, leaving 27 students to be reseated in new seats. The total number of possible combinations is 28C1 = 28. However, only one of these combinations will result in exactly one student being reseated in their original seat. Therefore, the probability of exactly one student being reseated in their original seat is 1/28.

To calculate the probability of at least one student being reseated in their original seat, we can use the complement rule. This means that we can calculate the probability of no students being reseated in their original seat and subtract it from 1. The probability of no students being reseated in their original seat can be calculated by taking the total number of possible combinations of 28 students being reseated in new seats, which is 28!, and dividing it by the total number of possible combinations of 28 students being reseated in any seats, which is 28^28. This gives us a probability of (28!/28^28). Subtracting this from 1 gives us a probability of at least one student being reseated in their original seat as 1 - (28!/28^28).

In conclusion, the probability of exactly one student being reseated in their original seat is 1/28 and the probability of at least one student being reseated in their original seat is 1 - (28!/28^28). I hope this helps with your analysis and understanding of the probabilities involved in this scenario.