In how many floors does the elevator have to stop?

Question

In a high-rise building with ten floors above the ground floor an elevator is installed. Suppose that twelve people enter the elevator on the ground floor and independently select one of the ten floors under uniform distribution. In how many floors does the elevator have to stop on average to let off one or more people?

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Do we model the facts as a probability space as follows?

$\Omega=\{(x_1, \ldots , x_{10}) \mid x_i\in \{0,1\}\}$

$x_i=\left\{\begin{matrix} 1, & \text{ stops } \\ 0, & \text{ doesn't stop } \end{matrix}\right.$

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To calculate in how many floors the elevator has to stop on average to let off one or more people, do we have to calculate the expectation? In this case we have discrete uniform distribution since we consider the number of floors and not all natural numbers, or?

So, we have to calculate the following: $$E(X)=\sum_{k=1}^{10}x_k\cdot P(X=x_k)$$ right? But how can we calculate $P(X=x_k)$ ?

Answer 1

I would define $Y_i$ as a Bernoulli random variable that takes the value $1$ if the elevaror does not stop on the $i$th floor, and $0$ otherwise. Now, $X=10-\sum_{i=1}^{10}Y_i$ represents the number of stops.

Now, $$\eqalign{\mathbb{E}(X)&=10-\sum_{i=1}^{10}\mathbb{E}(Y_i)\cr &=10-\sum_{i=1}^{10}\mathbb{P}(Y_i=1)}$$ The event $(Y_i=1)$ means that no body left the elevator on the $i$th floor, and this happens with probability $(9/10)^{12}$. So $$\mathbb{E}(X)=10\left(1-\left(\frac{9}{10}\right)^{12}\right)\approx7.176$$

Answer 2

Hint:

First, try to find out the average number of lifts used when $x[1,2,3,4,5,6,7,8,9,10,11,12]$ people get onto one of the $i[10]$ lifts... $$x = 1 ,i = 1 $$ $$x = 2, i = ?$$