0
$\begingroup$

$20$ people board a train they are equally likely to get off at either of the next $10$ stations and choose where to get of independently. Determine the expected number where no one gets off the train.

Here is what I got so far:

Define $X_i=1$ if nobody gets off and $X_i=0$ if at least one person gets off for $i=1,2,...10$. Define $X=\sum_{i=1}^{10} X_i$. Then we want $E[X]=E[X_1]+...+E[X_{10}]$ Now I feel like these are identical but I'm not sure so if they are then $E[X]=10E[X_1]$ as I said I'm not sure on this and even if it is true I don't know how to get $E[X_1]=1 \times P(\text{No one gets off})$.

Is this the right approach? And how do I complete this?

Thanks.

$\endgroup$
2
  • $\begingroup$ It is true that $E[X]=10E[X_1]$. As for the probability that noone gets off at stop one, Let $A_1,A_2,\dots,A_{20}$ be the events that person $1,2,\dots,20$ dont get off at stop $1$ respectively. Recognize then that $X_1=1$ if and only if we are in the event $A_1\cap A_2\cap\cdots \cap A_{20}$. What is $Pr(A_1\cap A_2\cap\cdots \cap A_{20})$? Remember the independence assumption. $\endgroup$
    – JMoravitz
    Apr 13, 2017 at 19:50
  • $\begingroup$ Is the final answer $1.216$? $\endgroup$
    – Ben B
    Apr 13, 2017 at 20:03

1 Answer 1

0
$\begingroup$

$$\mathbb{E}[X] = 10\mathbb{E}[X_1]$$

is correct. By independence, we have

\begin{align}\mathbb{E}[X_i]&=Pr(X_i=1) \\ &= Pr(\text{ no one gets off at } i) \\ &= \left( 1- \frac1{10} \right)^{20}\end{align}

$\endgroup$
2
  • $\begingroup$ So the final answer would be $10 \times (0.9)^{20} \approx 1.216$ $\endgroup$
    – Ben B
    Apr 13, 2017 at 20:02
  • $\begingroup$ yup, my answer is the same. $\endgroup$ Apr 13, 2017 at 20:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .