Train waiting time in probability 
Let's say a train arrives at a stop every 15 or 45 minutes with equal probability (1/2). What is the expected waiting time of a passenger for the next train if this passenger arrives at the stop at any random time. This means that the passenger has no sense of time nor know when the last train left and could enter the station at any point within the interval of 2 consecutive trains.

I was told 15 minutes was the wrong answer and my machine simulated answer is 18.75 minutes. I just don't know the mathematical approach for this problem and of course the exact true answer. Sincerely hope you guys can help me. 
 A: Picture in your mind's eye the whole train schedule is already generated; it looks like a line with marks on it, where the marks represent a train arriving. On average, two consecutive marks are fifteen minutes apart half the time, and 45 minutes apart half the time. 
Now, imagine a person arrives; this means randomly dropping a point somewhere on the line. What do you expect the distance to be between the person and the next mark? First, think of the relative probability of landing in each gap size of mark, and then deal with each case separately. 
Does this help? I can finish answering but I thought it's more available to provide some insight so you could finish it on your own. 
A: Your simulator is correct. Since 15 minutes and 45 minutes intervals are equally likely, you end up in a 15 minute interval in 25% of the time and in a 45 minute interval in 75% of the time.
In a 15 minute interval, you have to wait $15 \cdot \frac12 = 7.5$ minutes on average.
In a 45 minute interval, you have to wait $45 \cdot \frac12 = 22.5$ minutes on average.
This gives a expected waiting time of $\frac14 \cdot 7.5 + \frac34 \cdot 22.5 = 18.75$.
