1 elevator with 8 ppl stops at all of the remaining 11floors,what probability of no two ppl off thesamefloor?

Question

hard probability question,,the more detail, the better it is,,thanks

I think the denominator is 11^8,there is 11^8 ways

Answer 1

Well, there are two events: one big and one small and forms a small part of the first one.

The big is that the 8 people get from a combination of the 11 floors elevators.
The small is that each one gets off a different elevator from the rest.

big event can happen in 11^8 ways since each one can get off 11 elevators and they are independent such that no body's action will affect any of the others.

small event will happen in 11*10*9*...*4 = 11!/3! since
we can number them and then the first can get off 11 elevators, the second 10 remaining ones, the third 8, ...the last 4.

So the probabilit is that of the small event which is
(11!/3!)/11^8 = 0.031 approximately.

Answer 2

This is my take on the problem:
8 people, 11 floors. Each person has 1/11 chance to get off at any floor.
Person 1 has 11 floors to get off on = 11/11
Person 2 has 10 floors to get off on = 10/11
Person 3 has 9 floors to get off on = 9/11
Person 4 has 8 floors to get off on = 8/11
Person 5 has 7 floors to get off on = 7/11
Person 6 has 6 floors to get off on = 6/11
Person 7 has 5 floors to get off on = 5/11
Person 8 has 4 floors to get off on = 4/11
P(no 2 people on same floor) = (11!/3!)/11^8 = 0.031

Answer 3

The number of possible combinations where eight people get off the elevator on eight different floors is:

11C8 = 165.

Each of the eight people can exit on any of the eleven floors so the number of possible combinations is:

11^8

So the probability is:

(11C8) / 11^8

Answer 4

Who (what) gets off at the last floor?

..Remaining 11 floors?

Answer 5

if no two people means at most one person, then solution is:

The total no. of ways for leaving elevator for 8 persons at 11 floors is
8^11.(as any no. of persons can get down at the same floor.)
Now the favorable condition is to get down at most 1 person at a floor. so total no. of ways is 11C8.
Hence, the required probability is (11C8)/(8^11) = 165/8589934592
=1.92085E-8. Ans

if no two people means any no. of persons can get down except two at a floor, then solution is :

The total no. of ways for leaving elevator for 8 persons at 11 floors is
8^11.(as any no. of persons can get down at the same floor.)

Now the favorable condition is to get down any no. of persons except two at a floor. to find this first we find the no. of ways of getting down exactly 2 persons at a floor. so total no. of ways is {(11C1 x 8C2) x (10C1 x 6C2) x (9C1 x 4C2) x (8C1 x 2c2)}. here first we choose 1 floor from 11 floors and 2 persons from 8 persons.this will give 11C1 x 8C2 ways, then 2 persons from remaining 6 persons will be get down at one of remaining 10 floors and so on... this will give 19958400 ways.

So,the no. of ways of getting down any no. of persons except two at a floor = (8^11) - 19958400 = 8569976192
Hence, the required probability is = 8569976192/8589934592 = 0.99768.

Answer 6

There are many ways this can work, each with its own probability. Which one is it?