Question

The number of people who enter an elevator on the ground floor is a Poisson random variable with mean $10$. If there are $N$ floors above the ground floor, and if each person is equally likely to get off at any one of the$N$ floors, independently of where the others get off, compute the expected number of stops that the elevator will make before discharging all of its passengers.

Short answer

$E\left(X\right)=N-N{e}^{-\frac{10}{N}}$

Step-by-step solution

Given information

The number of people who enter an elevator on the ground floor is a Poisson random variable with mean $10$. If there are$N$ floors above the ground floor, and if each person is equally likely to get off at any one of the$N$ floors, independently of where the others get off, compute the expected number of stops that the elevator will make before discharging all of its passengers.

Explanation

Characterize arbitrary variable $X$as the quantity of floors that have been utilized in transport and characterize $N$as the quantity of individuals at the ground floor. We know that $N~Pois\left(10\right)$. Assuming we are given data that $N=n$, we can compose

where $I_k$is marker irregular variable which shows regardless of whether the elevator has halted on $k$th floor.

See that

The total expectation 

Since the elevator won't stop on the principal floor if and provided that every one individuals have picked a portion of the leftover floors. Utilizing the law of the total expectation, we have that

$E\left(X\right)=\sum _{n=0}^{\infty }E(X\mid N=n)P(N=n)$

$=\sum _{n=0}^{\infty }NE\left(I_1\right)P(N=n)$

$=\sum _{n=0}^{\infty }N\left[1-{\left(\frac{N-1}{N}\right)}^n\right]\frac{{10}^n}{n!}{e}^{-10}$

$=N-N{e}^{-10}\sum _{n=0}^{\infty }\frac{1}{n!}{\left(10\left(1-\frac{1}{N}\right)\right)}^n$

$=N-N{e}^{-\frac{10}{N}}$

Final answer

The expected number of stops that the elevator will make before discharging all of its passengers are,

$E\left(X\right)$$=N-N{e}^{-\frac{10}{N}}$