Math Help - Stuck on this for a while :(! Need help

1. Stuck on this for a while :(! Need help

there are 14 students in a class in it's first session. in the next session, a new student joins and was the first to enter the class and chose a random seat. the 14 other students entered the class one by one and each student took the seat he/she had in the last session if it was empty. If the seat was taken, the student picked an empty seat at random. if there are 15 seats in the class, what is the probability that the last student entering would sit on the same seat he/she had in the previous class? same question again but with 20 seats in the class....

2. Originally Posted by beckso22
there are 14 students in a class in it's first session. in the next session, a new student joins and was the first to enter the class and chose a random seat. the 14 other students entered the class one by one and each student took the seat he/she had in the last session if it was empty. If the seat was taken, the student picked an empty seat at random. if there are 15 seats in the class, what is the probability that the last student entering would sit on the same seat he/she had in the previous class? same question again but with 20 seats in the class....
The first part is rather simple.
Question, does the part in red refer to the same 15 students or do we change the question to 19 students? The second way is simple also. But to keep the number 15 is more difficult.

3. The number of students is still 15... Can you help out with the first part though? how would you go about solving that? Thanks

4. Suppose that the chairs are labeled A through O. Let’s say the absent student was assigned chair A, but does not know it. Further, suppose that the last student to come on the second day happens to be assigned chair O.
Now if when the last student arrives chair O is not occupied he takes. If it is taken he takes another. Therefore, chair O is occupied in any case. There are (15!) ways to have chairs occupied that day. All the ways that end in O mean the last student to arrive is in the correct seat. Therefore the answer is $\frac{14!} {15!} = \frac {1} {15}$

The model is much the same, but the answer is not as nice.
${{19} \choose {14}}(15!)$ is the number of ways that chair O must be occupied.

5. I get the first part, but the second one I still don't quiet understand...Thanks for the help

6. Look at the problem this way.
After the last student enters the room and seats him/herself, there are 15 seats occupied.
The seat assigned to that last to enter must be occupied.
So there is a one in fifteen chance that the correct person is seated there.

7. ok I get it now , but when the last student enters and there are 6 empty chairs left (the case where we start with 20 chairs and 15 students) not just one, what is the probability that this student would get the same chair as last time? would that be 1/20??

8. No. It is $\frac {1}{15}$.

9. Now I'm confused...so both cases have the same result?!!

The first time I gave the numbers so you could calculate the answer.
The next I gave a common sense approach.

Suppose that there are fifteen chairs in a room. They are numbered 1 to 15.
I randomly assign 15 students, you are among them, to the 15 chairs.
Do you agree that the probability that you will be in #15 is $\frac {1} {15}$?
I you do not, don’t continue to read this. Someone else will try to help.

Here is the point of the problem. Say on the first day someone was assigned #15 out of 20. For the sake of argument we suppose that person is the last to arrive for the second lecture. Fourteen others have entered before that student and have seated themselves according to the rules. When this last person enters the hall if #15 is available the person must take it or else find another seat. In any case, seat #15 is occupied and is among only fifteen occupied seats. What is the probability that the last person is in seat #15, the correct seat?

11. Do you agree that the probability that you will be in #15 is 1/15? I agree

as for the second question, I still don't see what the solution would be, but listen don't get frustrated, I appreciate your help .

12. The answer is not 1/15. Maybe that poster read the question incorrectly. The students don't choose chairs at random. If their seat from the previous session is available they take it, otherwise they choose at random from the remaining seats.

The answer is surprisingly 1/2. Its a bit lengthy to explain why. It comes down to two possibilities: Which chair gets taken first, the last student's or the empty seat. If the empty seat is taken first, then the last student gets his chair for sure. Obviously if his chair is taken before the empty seat, then he doesn't get it.

In the case of 20 chairs. Now you have 6 empty seats and so the probability is 6/7.

I know this explanation is not very clear. Let me know if you still need help with this and I'll try to be a bit more clear.