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]

hi! i think the answer is 1/2 : )

sas (side angle side)

aa (angle angle)

sss(side side side)

sas find two sides with an angle in between

aa is finding two angles and the third will add up to 180

sss find all sides

step-by-step explanation: