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Maths Problem

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B1rd
2,974 posts
Joined December 2013
Topic Starter
B1rd
Maybe someone who is good at maths can tell me the answer to this problem.

There are two people who will meet at a specified place and time, with each arriving at a random time somewhere within a 60 minutes period. The first person who arrives will wait 15 minutes before leaving if the other person hasn't shown up yet. What is the chance they will meet?
johnmedina999
296 posts
Joined October 2016
johnmedina999
My first and most obvious guess is 1/4 chance, but that's not the answer, is it?
ColdTooth
1,100 posts
Joined June 2012
ColdTooth
1/60th
B1rd
2,974 posts
Joined December 2013
Topic Starter
B1rd
Put some actual thought in to in and explain your reasoning.
johnmedina999
296 posts
Joined October 2016
johnmedina999

B1rd wrote:

Put some actual thought in to in and explain your reasoning.
That's what I feared.

After the first person gets to the destination, the other person has 15 minutes to arrive. 1/4 would work if the first person got to the destination exactly when the 60 minute window opened. But what if he arrives five minutes after the window opens? Then the probability is 15/55. Ten minutes? 15/50. Fifteen minutes after? 15/45. From this pattern, we can create the formula 15/(60-x), where x is the number of minutes the first person took to get to the destination.

Is this the right answer?
abraker
Global Moderator
8,295 posts
Joined July 2014
abraker

B1rd wrote:

There are two people who will meet at a specified place and time, with each arriving at a random time somewhere within a 60 minutes period. The first person who arrives will wait 15 minutes before leaving if the other person hasn't shown up yet. What is the chance they will meet?
Answer is 5/9. Explanation here: https://math.stackexchange.com/question ... g-in-a-bar
johnmedina999
296 posts
Joined October 2016
johnmedina999

abraker wrote:

Answer is 5/9. Explanation here: https://math.stackexchange.com/question ... g-in-a-bar
These two problems are different:

Two people have to spend exactly 15 consecutive minutes in a bar
In the problem linked, the two cannot arrive fifteen minutes before time is up, where here they can. The linked problem doesn't take into account the possibility of both meeting in the last five minutes, while this one does. Your answer is most probably wrong.

(If 5/9 really is the answer I will be mad)
abraker
Global Moderator
8,295 posts
Joined July 2014
abraker
Then it will be:

Box area:
60x60 = 3600

Two triangle area:
2×1/2×45×45=2025

Probability:
(3600−2025)/3600=1125/2025=43.75%


If that's still not it, just use the following formula for whatever values it is:
[(total_time_period^2 - (total_time_period - wait_period)^2)]/total_time_period^2
johnmedina999
296 posts
Joined October 2016
johnmedina999
Yes, that's the answer, I was doing it as you typed that.

I did it graphically (because I have no access to paper currently).





There's your answer, B1rd.
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