Two friends who take the metro to their jobs from the same station arrive to the station uniformly randomly between 7:00 and 7:20 in the morning. They are willing to wait for one another for 5 minutes, after which they take a train whether together or alone. What is the probability of their meeting at the station?
Hello, i just can't get a hang of such problems. There is supposed to be two methods, graphical and integration. Is this method correct of integration:
Probaility= $\int_{0}^{20}\int_{x-5}^{20} \frac{1}{400}dy dx + \int_{0}^{20}\int_{0}^{x+5} \frac{1}{400}dy dx $
And what could be the graphical methid? Any help will be highly appreciated.
Your integrals are substantially overcounting the situations where they encounter each other, as well as including some situations where they shouldn't meet up at all. You can set it up in that fashion, but one integral should cover those situations where one friend arrives up to five minutes earlier than the other, and the other integral should cover the converse.
Actually, because you also have to take care of boundary situations, because neither friend ever shows up before 7:00. So, you'll end up with
$$ \int_{x=0}^5 \int_{y=0}^x \frac{1}{400} \, dy \, dx + \int_{x=5}^{20} \int_{y=x-5}^x \frac{1}{400} \, dy \, dx + \\ \int_{y=0}^5 \int_{x=0}^y \frac{1}{400} \, dy \, dx + \int_{y=5}^{20} \int_{x=y-5}^y \frac{1}{400} \, dy \, dx $$
It's a bit tedious, but should arrive at the proper answer.
The graphical method is to sketch out a square labelled 7:00 to 7:20 on the $x$-axis, and 7:00 to 7:20 on the $y$-axis. Each point in this square represents a time of arrival for one friend, and another time of arrival for the other friend. Shade in the area that represents situations where the friends meet. The fraction of the square that is shaded in is the probability you want.