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.

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 $a = b$.
