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.
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
\[\frac{1100}{3600} = \boxed{\frac{11}{36}}.\]
11/36. Hope this helps.
I personally believe that giving a hint gets one started on a great start.
We should graph this problem! Try defining the start and end times and see where you get. Can you use those definitions to find where the two people meet?
The picture in this problem really plays a big part in solving this problem. Once you draw it, do you see what (x, y) makes? Can you find the area of this figure?
Also, to maintain appropriate comments and a fun learning environment, I wouldn't go off offending others when it's not your personal problem. I'm sure AoPS is aware of some people plagiarizing their solutions/questions, but we are all simply trying to learn from AoPS. So please don't put hurtful comments!
just saying, you shouldn't criticize people because they got the answer wrong.
but yes, it is 11/36
@Michael and @mathfailure... I think that you BOTH copied off this page...
artofproblemsolving.com/class/1969-mathcounts-advanced/homework/8
you have committed a crime...
@mathfailure, you copied the last problem. At the very bottom of the page, it says: Copyright © AoPS Incorporated. This page is copyrighted material. You can view and print this page for your own use, but you cannot share the contents of this page with others.
@Michael, you copied the answer to the last problem, and right below it at the very bottom of the page, it says: Copyright © AoPS Incorporated. This page is copyrighted material. You can view and print this page for your own use, but you cannot share the contents of this page with others.
@Michael and @mathfailure, you have just committed plagiarism and have broken the law. >: (
I can also see this because you forgot to get rid of signs of LATEX because you copied directly(kind of a formatting code).
Also, I agree with @I AIN'T GOT NO FIRST NAME.
PS:
Other people, you cant see this AOPS page because you need to sign up and pay for AOPS
oh yes of course its 11/36
...8 years later
we get it you guys
no need for more solutions
EXPOSED LOOL
@TATLLE TALER LOOOOL
wat