# Probability of catching subway.

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?

• 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. – Jon Claus May 20 '13 at 2:22
• @JonClaus I am comfortable with calculus, not so much with statistics. Would it be two integrals then for the two different cases? – Joakim May 20 '13 at 2:25

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 $\frac{1}{6}$.
• By the way, was the answer $\frac{11}{72}$ of $\frac{1}{6}$? – Jon Claus May 20 '13 at 3:13