In how many floors does the elevator have to stop?

[mathmari]

Hey!

I am looking at the following:
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?
I have done the following:

Let $i\in \{1, \ldots , 10\}$.

Let $X_i$ be a random variable that takes the value $1$ if at least one person gets off on floor $i$ and $0$ otherwise.

Then we get $X_i=\left\{\begin{matrix}
1, & \text{ stops on floor } i\\
0, & \text{ otherwise }
\end{matrix}\right.$ We want to calculate the expectation of the number of stops, right?

The probability that nobody gets out on floor $i$ is equal to
\begin{align*}P&(\text{Nobody gets out on floor } i ) \\ & =P\left ((\text{Person } 1 \text{ doesn't get out}) \cap (\text{Person } 2 \text{ doesn't get out}) \cap \ldots \cap (\text{Person } 12 \text{ doesn't get out})\right ) \\ & = P\left (\cap_{j=1}^{12}(\text{Person } j \text{ doesn't get out})\right )\end{align*}

SInce each person chooses the floor to get out independently from the others, it holds that
\begin{equation*}P\left (\cap_{j=1}^{12}(\text{Person } j \text{ doesn't get out})\right )=\prod_{j=1}^{12} P(\text{Person } j \text{ doesn't get out})\end{equation*}

The probability that a person gets out on floor $i$ is equal to $\frac{1}{10}$. The probability that the person doesn't get out on floor $i$ is equal to $1-\frac{1}{10}=\frac{9}{10}$.

We get the following:
\begin{equation*}\prod_{j=1}^{12} P(\text{Person } j \text{ doesn't get out})=\prod_{j=1}^{12}\frac{9}{10}=\left (\frac{9}{10}\right )^{12} \end{equation*} Is everything correct so far? (Wondering) Which is the formula for the expectation that we have to use? Do we maybe have to use the formula $E(X)=\sum_{k=1}^{10}x_k\cdot P(X=x_k)$ ? (Wondering)

 

[I like Serena]

Hi mathmari! (Smile)

I believe that is correct yes.

So we have:
$$P(\text{elevator does not stop at floor }i) = \left (\frac{9}{10}\right )^{12}$$
Either the elevator stops or not at floor $i$, so the expected number of stops on floor $i$ is:
$$E(\text{nr of stops on floor }i) = 0 \cdot \left (\frac{9}{10}\right )^{12} + 1 \cdot \left(1 - \left (\frac{9}{10}\right )^{12}\right)$$
Thus the expected total number of stops is:
$$E(\text{nr of stops}) = \sum_{i=1}^{10}E(\text{nr of stops on floor }i) = 10 \cdot \left(1 - \left (\frac{9}{10}\right )^{12}\right)$$
(Thinking)

 

[mathmari]

Either the elevator stops or not at floor $i$, so the expected number of stops on floor $i$ is:
$$E(\text{nr of stops on floor }i) = 0 \cdot \left (\frac{9}{10}\right )^{12} + 1 \cdot \left(1 - \left (\frac{9}{10}\right )^{12}\right)$$

Could you explain to me how we get that? (Wondering)

 

[I like Serena]

Could you explain to me how we get that? (Wondering)

Generally the expectation of a random variable X is:
$$EX=\sum x_k P(x_k)$$
where the $x_k$ represent the possible outcomes.

In our case the possible outcomes for a specific floor i are 0 stops and 1 stop. (Thinking)

 

[mathmari]

Generally the expectation of a random variable X is:
$$EX=\sum x_k P(x_k)$$
where the $x_k$ represent the possible outcomes.

In our case the possible outcomes for a specific floor i are 0 stops and 1 stop. (Thinking)

I understand! Thank you very much! (Mmm)