Probability theory question about trains and it schedule

Question

I received some question for my exam preparation, but I'm not sure about the answer. So I need some reasonable explanation of the following tasks:

I. Every day Sonja arrives at the railway station between 6 and 7 pm, but her exact time of arrival is random within this one hour interval. The trains leave at 6.30 and 7.00: what is the probability that Sonja has to wait more than 10 minutes?

As I suppose, the answer is 2/7 (i.e., 00. 10, 20, 30, 40, 50, 60 in one hour and train leaves at 30 and 00 so we just relate these two numbers). But I think that this one is incorrect.

Please, explain me the meaning of the task and how to calculate the result.

Answer 1

Sonja can arrive anywhere from $6:00-6:20 $ and from $6:30-6:50$. These times result in her having to wait for more than $10$ minutes. This is $40$ minutes out of $60$, so the probability that Sonja has to wait for more than $10$ minutes is $2/3$

Answer 2

Let $t$ be the time in hours after 6pm that Sonia arrives. We have $t \in [0,1]$. Her waiting time will be $w(t) = \begin{cases} {1 \over 2}-t, & t \le {1 \over 2} \\ 1-t, & t > {1 \over 2 } \end{cases}$.

Then the set $\{ t \in [0,1]| w(t) \ge {1 \over 6} \} = [0,{1 \over 3}] \cup ({1 \over 2}, {1 \over 2}+{1 \over 3}]$. Since her arrival time is uniform in $[0,1]$ we see that the probability of waiting 10 minutes or more is ${1 \over 3} + {1 \over 3} = {2 \over 3}$.