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?

  • $\begingroup$ If it a uniform distribution, then you cannot simply use discrete cases. Instead, you need an integral (as opposed to a summation). You might think this is too complex for your level (I don't know if you've taken calculus) but an integral represents area. A geometrical interpretation will help here. $\endgroup$ – Jon Claus May 20 '13 at 2:22
  • $\begingroup$ @JonClaus I am comfortable with calculus, not so much with statistics. Would it be two integrals then for the two different cases? $\endgroup$ – Joakim May 20 '13 at 2:25

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} $.

  • $\begingroup$ I drew it out and it made perfect sense! Thanks much! $\endgroup$ – Joakim May 20 '13 at 3:05
  • $\begingroup$ By the way, was the answer $ \frac{11}{72} $ of $ \frac{1}{6} $? $\endgroup$ – Jon Claus May 20 '13 at 3:13
  • $\begingroup$ Even if it was randomly constant, is there not still the new hour threshold. To me 11/72 makes the most sense. The problem states randomly chosen time within any hour. $\endgroup$ – Joakim May 20 '13 at 3:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.