2

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?

$$$$

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

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

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

$$$$

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 $P(X=x_k)$ ?

Mary Star
  • 13,956
  • 1
    I think by $P(X=x_k)$ you actually mean $P(X_k=1)$ ? anyway that formula wouldn't work because your $X_i$ are not independant: If you know that you don't stop at the second floor, the probability of stopping at the third floor is slightly higher. – Evargalo Jul 05 '17 at 15:29
  • 1
    I would go with defining $(X=i)$ as the event "the first stop of the elevator is the floor i". – krirkrirk Jul 05 '17 at 15:33
  • 1
    See this answer. – Fabio Somenzi Jul 05 '17 at 15:41
  • 1
    Possible duplicate of https://math.stackexchange.com/q/214713/265466, the top related question at right. – amd Jul 05 '17 at 21:01

2 Answers2

2

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 $$\mathbb{E}(X)=10\left(1-\left(\frac{9}{10}\right)^{12}\right)\approx7.176$$

Felix Klein
  • 1,871
  • Why is it better to define $Y_i$ as you did and not as I did the $x_i$ ? Also how did you calculated $X$ ? – Mary Star Jul 05 '17 at 21:45
  • 1
    @MaryStar, Because the probability of the event $(Y_i=1)$ is easy to calculate, this happens when the each person does not choose the $i$th floor to get out, and for each person this happens with probability 9/10. Now, we have 10 floors and when we subtract the number of floors where the elevator does not stop it remains the number of floors where it stops. This is $X$. – Felix Klein Jul 06 '17 at 03:39
  • the probability that the elevator stops on floor i is 1/10. therefore the probability that the elevator does not stop there is 1-1/10=9/10. But how do we get the exponent 12? Could you explain it to me? I got stuck right now. – Mary Star Jul 06 '17 at 04:07
  • 1
    @MaryStar, Yes, each person chooses not to stop with probability $9/10$ independently of the others, and there are $12$ of them. – Felix Klein Jul 06 '17 at 05:02
  • Ah ok! Did you use the formula of $\mathbb{E}(X)$ that I wrote in my initia lpost? Or which is the general formula? – Mary Star Jul 06 '17 at 07:27
  • 1
    @MaryStar, I used the linearity of the expectation. – Felix Klein Jul 06 '17 at 16:48
  • Ah ok! Thank you very much!! :-) – Mary Star Jul 06 '17 at 17:11
1

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... $$x = 1 ,i = 1 $$ $$x = 2, i = ?$$

Xetrov
  • 2,089