probability question  help!
posted by mathfailure on .
Two people agree to meet at a coffee shop. They each independently pick a random moment in time between 8 a.m. and 9 a.m. and show up exactly at their selected time. But they are very impatient, and only stay for 10 minutes after when they arrive. What is the probability that they meet? Express your answer as a common fraction.

you ask two many question but the answer is 1/6

Gee you guys always seem to provide wrong answers to easy problems. By using geometric probability, we find the answer to be 11/36.

11/36

mathcounts and yea are correct
We solve the problem by graphing. Let $x$ and $y$ be the arrival times of the two people in minutes after 8 a.m., so $0 \le x,y \le 60$. Then the two people meet if and only if $x  y \le 10$. We graph the set of points $(x,y)$ such that $x  y \le 10$.
[asy]
unitsize(0.08 cm);
filldraw((0,0)(10,0)(60,50)(60,60)(50,60)(0,10)cycle,gray(0.7),invisible);
draw((0,0)(60,0)(60,60)(0,60)cycle);
draw((10,0)(60,50));
draw((0,10)(50,60));
label("$x$", (60,0), E);
label("$y$", (0,60), N);
dot("$(0,0)$", (0,0), SW);
dot("$(60,0)$", (60,0), S);
dot("$(60,60)$", (60,60), NE);
dot("$(0,60)$", (0,60), W);
dot("$(10,0)$", (10,0), S);
dot("$(60,50)$", (60,50), E);
dot("$(50,60)$", (50,60), N);
dot("$(0,10)$", (0,10), W);
[/asy]
The set of all points $(x,y)$ is a square with area $60^2 = 3600$.
The area outside the "successful" region consists of two right, isosceles triangles, with legs 50 and 50. The area of each triangle is $1/2 \cdot 50 \cdot 50 = 1250$, so the area of the "successful" region is $3600  2 \cdot 1250 = 1100$. Therefore, the probability that the two people meet is
\[\frac{1100}{3600} = \boxed{\frac{11}{36}}.\]