Probability: random elevator stops I have a elevator probability problem but with a twist (instead of people exiting the elevator, it is the elevator stopping on each level). Require some help understanding and completing the probability questions.
Problem:
4 people go into the elevator of a 5-storey shophouse. Assume that each of them exits the building uniformly at random at any of the 5 levels and independently of each other. N is the random variable which is the total number of elevator stops.


*

*Describe the sample space for this random process.


Answer: {1,2,3,4,5}
Understanding: Sample space represents all the possible outcomes of the random event. Hence, the elevator can only stop at these 5 floors. Just like throwing a die and randomly getting a number, the sample space for that is {1,2,3,4,5,6}.


*Let $X_i$ be the random variable that equals to 1 if the elevator stops at floor i and 0, otherwise. Find the probability that the elevator stops at both floors $a$ and $b$ for $a$, $b$ $\in$ {1,2,3,4,5}. Find $EX_aX_b$. 


Question: Do they mean the elevator stop at two consecutive floors or two different floors? Can a and b be the same, a smaller than b and vice versa?


*Prove the independence of $EX_1X_2$.

*Calculate $EN^2$ (represented as ($X_1 + ...+ X_5)^2 = \sum_{(i,j)}  X_aX_b$ where the sum is over all ordered pairs (a,b) of numbers from {1,2,3,4,5} and the linearity of expectation. Find the variance of N.

*Determine the distribution of N. That is, determine the probabilities of events N = i. Compute $EN and EN^2$ directly by using the laws of expectation.
I got stuck from question 2 onwards with thinking of the possible ways that this can be done. It would be superb if someone can shed some light on them. Many thanks.
 A: You did not specify the events $X_a$ and $X_b$, but from the context I assume it is meant that
$$ X_a = 1\{\text{elevator stops at floor } a\}. $$
You have to find $\mathbb{E}X_aX_b$, which is equal to the probability that the elevator stops at both floor $a$ and $b$. In the question it is only stated that $a,b \in \{1,\dots,5\}$ but it is not specified that they can't be equal, meaning we should also answer the question in the event $a = b$. Moreover, it does not have to be the case that $a$ and $b$ are two consecutive floors, since in that case they would have stated that $a \in \{1,\dots,5\}$ and $b = a+1$. So you should just read this question as: given two floors, what is the probability that the elevator stops at these two floors (or, in case $a = b$: given this floor, what is the probability that the elevator stops at this floor).
As mentioned above, we have two different scenario's: $a$ and $b$ different floors or $a = b$ and we only consider one floor. I will now give a hint on answering the question for both scenario's:
$a = b$:
in the case $a = b$ we have to find $\mathbb{P}(\text{ elevator stops at floor a})$. Since the only way the elevator stops at this floor is if there are people exiting the elevator at this floor, we have that:
\begin{align*}
\mathbb{P}(\text{elevator stops at floor a}) &= 
\mathbb{P}(\geq 1 \text{ people exit the elevator at floor a}) \\
&= 1 - \mathbb{P}(\text{no people exit at floor a}) 
\end{align*}
and we can compute this last probability, working with binomial random variables.
$a \neq b$:
in the case $a \neq b$ we have to find $\mathbb{P}(\text{ elevator stops at both floor a and b})$. A hint to find this probability is to use the rule: $\mathbb{P}(A \cap B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cup B)$ and use the results from the case $a = b$.
