(a) A building has n floors numbered 1,2,...,n, plus a ground floor G. At the ground floor, m people get on the elevator together, and each gets off at a uniformly random one of the n floors (independently of everybody else). What is the expected number of floors the elevator stops at (not counting the ground floor)

The expected number of floors the elevator stops at, not counting the ground floor is =

n*(1-(1-1/n)^m)

Step-by-step explanation:

Here, we want to know the expected number of floors the elevator stops at.

let X1,X2,X3,..Xn are indicator variable for which value =1 if at least one person stops on that floor otherwise value is 0

P(at least one person stops at floor Xj)=1-P(none of m people stops at floor j)

=1-(1-1/n)^m

here total number of floors on elevetor Stops X=X1+X2+X3+...+Xn

hence expected number of floors on elevetor Stops

E(X)=E(X1)+E(X2)+E(X3)...+E(Xn)

=(1-(1-1/n)^m )+(1-(1-1/n)^m )+(1-(1-1/n)^m )+(1-(1-1/n)^m )+ n times

=n*(1-(1-1/n)^m)

the discriminant is 64 + 4*1*12 = 112

this is prime.

step-by-step explanation:

i only know the sequence of arithmetic it adds the first is 5+4 then it goes up as it adds to then it is 9+5 then 14+6 then to 20+7 and so on. 5,9,14,20,27,35.

step-by-step explanation: