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?
I don't know Asymptote or any other graphics rendering language, so you're out of luck with that. I'll do my best to describe what the situation looks like.
Let the $x$-axis vary from $ 0 $ to $ 60 $ to denote the time of the hour at which the man arrives. Note that if the man arrives at time $ x $, then the tram must arrive at time $ t $ that satisfies $ x \le t \le t + 10 $, for the most part.
Now, plot let the $y$-axis denote $ t $. Your area of interest will measure $ 60 \times 60 $. For $ 0 \le x \le 50 $, the interval of $ t $ satisfying inequality has width $ 10 $. Hence, the area of interest for $ x \le 50 $ is $ 50 \cdot 10 $; to be precise, it is a parallelogram bounded by the equations $ x = 0 $, $ x = t, x = t + 10, x = 50 $. For the interval $ 50 \le x \le 60 $, the area of interest is a triangle; it is bounded by $ x = 50, t = 60, x = t $. It has area $ 50 $. Hence, the probability is $ \frac{500 + 50}{60 \cdot 60} = \frac{11}{72} $.
This assumes that if the man arrives at, for example, 11:50, then the bus tram cannot arrive at 12:00 to pick him up because that is in a different hourly interval. If the tram comes once per hour at random constantly, then the probability is simply $ \frac{1}{6} $.