Brain Teasers
Brain Teasers Trivia Mentalrobics Games Community
Personal Links
Submit a Teaser
Your Favorites
Your Watchlist
Browse Teasers
All

Cryptography
Group
Language
Letter-Equations
Logic
Logic-Grid
Math
Mystery
Optical-Illusions
Other
Probability
Rebus
Riddle
Science
Series
Situation
Trick
Trivia

Random
Daily Teasers
Search Teasers

Advanced Search
Add to Google Add to del.icio.us

More ways to get Braingle...

Walk in the Park

Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

 

Puzzle ID:#21639
Fun:*** (2.47)
Difficulty:*** (2.69)
Category:Probability
Submitted By:Dedrik**
Corrected By:MarcM1098

 

 

 



Alice and Bob agree to meet in the park. They agree to each show up some random time between 12:00 PM and 1:00 PM. Wait 20 minutes for the other person, or until one o'clock, whichever comes first.

Assuming they stick to their word, what is the probability that they will meet?





What Next?

  
  

See another brain teaser just like this one...

Or, just get a random brain teaser

If you become a registered user you can vote on this brain teaser, keep track of
which ones you have seen, and even make your own.

 



Comments

katiebeth123*
Mar 21, 2005

OMG SHUT UP
krishnanAin*
Mar 23, 2005

Nice teaser.
mschup
Mar 24, 2005

i have a headache.
psychedelictoad**
Mar 28, 2005

That teaser hurt my soul.
runscapekiller**
Mar 31, 2005

is an anser soposed 2 b that long
Dedrik
Apr 06, 2005

I know, but it was the only way I could think to explain it. Believeme I am open to a shorter explination
ANgelStARAus*
Apr 10, 2005

ok, i did not get that!
monkeygirl***
Apr 11, 2005

confusing...
cehnehdeh
Apr 17, 2005

WAY to complicated and purely math no fun!
pi202**
Apr 19, 2005

Really nice teaser. A good bit of hard thought never hurt anyone. Well done!
duluoz_jack
Apr 22, 2005

If they have both agreed to wait until 10 then the probability is 100% that they will meet. Or what am I missing?
jinzcarmela**
Apr 29, 2005

What the heck was that? I almost got my brains lost
brianz**
Apr 30, 2005

Nice teaser.
BrownEyesAus*
Jun 16, 2005

hard, HARD, HARD!!!
babygrl329**
Jun 24, 2005

confusing
bjoel4evr
Jul 25, 2005

Nice one. I think your answer explained the teaser very well.
cazjackrabbit
Aug 24, 2005

But couldn't the waiting time have been like 12.37-12.57? Ok... well even if I completely missed that it was still HARD!!! But it was good in a way. If you have hours to do it in- or you could just do it till one o'clock, whichever comes first. Ok that was a lame attempt at a joke.
shoulder_angelAus*
Sep 22, 2005

I didn't get your explanation... I guess that I should just stop reading probability teasers 'cause they make me feel dumb . Oh, well
shoulder_angelAus*
Sep 22, 2005

Oh... and don't be such a know-it-all bjoel4evr!
MissleMan27Aus
Jan 01, 2006

I didn't understand any of that!
mr_brainiac*
Jan 04, 2006

I agree with the Missileman. Deceptively confusing puzzle looks easy until you jump in.
kayleeskittiesAus*
Jan 21, 2006

Good teaser, boggled my brain there for a moment but I got it!
paul726Aus*
Feb 11, 2006

Excellent job. Well thought out!
(user deleted)
Apr 27, 2006

This riddle has a flaw. You can't assume that they will show up at 120. 12:20 or 12:40 unless you put that in the problem itself... any random time means any random time. It is impossible to solve otherwise.
Dedrik
Aug 09, 2006

I didn't assume she showed up at any of those times. I suggested breaking the hour up into different pieces as the probability of meeting up is different during each interval.
javaguru*us*
Dec 09, 2008

I approached the problem slightly differently. I broke the time period into the three 20 minutes sections. There are then three relevant possibilities:

1) They arrive in the same period. They always meet in these cases.
2) They arrive in non-adjacent periods. They never meet in these cases.
3) The arrive in adjacent periods. As explained above, there is a 50% chance they meet in these cases.

There are 3 x 3 = 9 combinations of periods they can arrive in.

There are 3 combinations where they both arrive in the same period: {(1,1), (2,2), (3,3)}

There are two combinations where they arrive in non-adjacent periods: {(1,3), (3,1)}

That leaves four combinations where they arrive in adjacent periods {(1,2), (2,1), (2,3), (3,2)}.

Therefore there is a:

3/9 x 1 + 2/9 x 0 + 4/9 * .5 = 5/9

probability of meeting.
opqpop
Sep 28, 2010

This is a standard geometric probability problem. Let the x axis be Alice's arrival time and y axis be Bob's arrival time. We're interested in the region inside a 60minute * 60minute square that corresponds to the event that they meet.

Notice this region is bounded by the line y=x shifted 20minutes up and 20 minutes down. This is because whatever time Alice arrives, Bob needs to be within 20 minutes of her time for them to meet.

The area of this region is 60 * 60 - 40 * 40. 40 * 40 is the area of the sum of the two triangles that aren't in this region. You'll see it clearly if you draw the graph.

The total area is 60 * 60. Hence (60-40)(60+40) = 2000 / 3600, so the answer is 5/9.



Back to Top
   



Users in Chat : Lex624, Willacora 

Online Now: 15 users and 657 guests

Copyright © 1999-2014 | Updates | FAQ | RSS | Widgets | Links | Green | Subscribe | Contact | Privacy | Conditions | Advertise

Custom Search





Sign In A Create a free account