A group of n people enter an elevator in a building with k floors. Each person independently selects a floor uniformly at random (each with equal probability 1/k). The elevator stops at each floor selected (by at least one person). Let N be the number of stops. (a) [3 marks] Find a simple expression for EN, the average number of stops. Bonus [2 marks] Suppose that n

Hence the person stop at floor by at least one person will be

E(X)=(summation from K=1 to k)[1-{(k-1)/k}^n]

Step-by-step explanation:

Given:

There are n peoples and k floors in a building.

Selects floor with 1/k probability .

To find :

Elevator stop at each floor by at least one person.

Solution:

Now

let K= number of floor at which at least one person will be stopping.

For getting E(X)

consider a variable Ak =1 if a least one person get of the elevator

and values for k=1,2,3.....k

K=(summation From k=1 to k)Ak

E(K)=((summation From k=1 to k) E[Ak]

=(summation From k=1 to k)[

Hence the person stop at floor by at least one person will be

E(K)=(summation from K=1 to k)[1-{(k-1)/k}^n]

congruent segments

step-by-step explanation:

da freeck kinda math is this just drop out

step-by-step explanation: