A blue tram shows up randomly in a uniform distribution given any hour of the day at a certain stop. A person shows up independently within this same hour. If they are only willing to wait 10 minutes maximum, what is the probability the tram will show up?
Attempt: If the tram shows up at say 5:01, the probability is $1/60$. Similarly, 5:02 would have probability $2/60$. This trend continues until the 11 minute mark where it would remain at a probability of $10/60$. If the tram stops at any time between 5:10 and 6:00 the probability will remain $10/60$.
My answer: $(5*9+10*51)/60*60$ or $\Pr = .1542$
Questions: First off, is this intuition right. Secondly, is there some kind of theory behind this problem to simplify it?