1

I am attempting to solve the following problem involving uniform distributions: $$ \\ $$

Two friends, Paul and Coralie, decide to meet sometime between 3:00pm and 4:00pm. Paul arrives at a time uniformly distributed between 3:15pm and 3:45pm whereas, Coralie arrives at a time uniformly distributed between 3:00pm and 4:00pm.

Part A: What is the probability that the friend that arrives first will wait less than 5 minutes for the other friend to join?

Attempt: I believe we would have to compute two cases here - one where Coralie has to wait less than 5 minutes and another where Paul has to wait less than 5 minutes: Here is a geometrical approach that I have considered:Geometric Approach Where I would have to compute the area of this figure to account for the times in-between each point (i.e. waiting for 4 minutes, 3 minutes and 20 seconds, etc.).

<p>The green points denote the times in which Coralie can arrive and waits 5 minutes for Paul [i.e. (Coralie Arrives, Paul Arrives), (3:15, 3:20), (3:20, 3:25),...(3:45, 3:50)]. </p>

<p>The blue points denote the times in which Paul can arrive and waits 5 minutes for Coralie [i.e. (Paul Arrives, Coralie Arrives), (3:10, 3:15), (3:15, 3:20),...(3:40, 3:45)].  </p>

<p>We have 6 squares and 12 triangles so the area is computed to be: <span class="math-container">$$Area= 6(\frac{5}{60}*\frac{5}{60})+12[\frac{1}{2}(\frac{5}{60}*\frac{5}{60})]= \frac{1}{12}=0.08333...$$</span>
The desired probability <span class="math-container">$=\frac{\frac{1}{12}}{\frac{60}{60}*\frac{30}{60}}=\frac{1}{6}=0.1666...$</span></p>

Part B: What is the probability that Paul arrives first?

Attempt: Using the same geometric approach as before, I have:

<p><a href="https://i.sstatic.net/9rGrO.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/9rGrO.png" alt="Geometric Approach 2"></a></p>

<p>The area of this figure: <span class="math-container">$$Area= [\frac{15}{60}*\frac{30}{60}]+[\frac{1}{2}(\frac{30}{60}*\frac{30}{60})]=\frac{1}{4}$$</span>
The desired probability <span class="math-container">$=\frac{\frac{1}{4}}{\frac{60}{60}*\frac{30}{60}}=\frac{1}{2}=0.5$</span></p>

Before doing any computations, I would like to confirm that I am heading in the right direction because I have a feeling that I am over-complicating it. Any suggestions would be appreciated!

  • For part A, you have computed the area for the numerator. Now divide by the area for the denominator. – RobPratt Apr 26 '20 at 17:16
  • I have the following: $\frac{\frac{1}{12}}{\frac{40}{60}*\frac{30}{60}}=\frac{1}{4}$; $\frac{40}{60}$ because along the y-axis we are going from 3:10-3:50 and $\frac{30}{60}$ because along the x-axis we are going from 3:15-3:45. –  Apr 26 '20 at 17:42
  • Why 40/60 instead of 60/60? – RobPratt Apr 26 '20 at 17:43
  • I just realized, I am using the time that Paul is available - which is 60/60 and the time Coralie is available - which is 30/60. –  Apr 26 '20 at 17:46
  • Related: https://math.stackexchange.com/questions/103015/chance-of-meeting-in-a-bar – Henry Mar 15 '25 at 07:57

1 Answers1

0

Hint: think geometrically. Each of the desired probabilities is a ratio of areas. The denominator is the area of a rectangle with side lengths $0.5$ and $1$.

RobPratt
  • 50,938
  • I initially attempted making a dotted grid (I added it above) - where the x- and y- axis start at 3:00pm, increasing in 5-minute increments, and ends at 4:00pm. For the first scenario, where Coralie waits less than 5 minutes, I have marked 7 points (3:15,3:20), (3:20, 3:25),...,(3:45, 3:50). I interpreted each of these probabilities as being $\frac{1}{144}$ because we have a 12 x 12 grid, but that would only account for these specific times. Would it be a sound idea to essentially calculate the areas of triangles as pictured? –  Apr 25 '20 at 20:08
  • That is the right idea, but the shape of interest will be one quadrilateral that contains the triangles you have already drawn. For each of Coralie’s arrival times, go up or down 5 minutes for Paul’s arrival times. – RobPratt Apr 25 '20 at 21:29
  • I made an adjustment to the grid, adding the points in which Paul can arrive and wait 5 minutes for Coralie. If I go up 5 minutes for Paul's arrival time - wouldn't that leave Coralie waiting 10 minutes? For example, for the point (3:15, 3:20) - if I increase Paul's arrive time by 5 minutes we would have (3:15, 3:25), i.e. a wait time of 10 minutes for Coralie. –  Apr 25 '20 at 23:40
  • Your grid now shows the quadrilateral I had in mind. The points correctly represent all pairs of arrival times that differ by at most 5 minutes. Now compute the area. – RobPratt Apr 26 '20 at 00:08