# Geometric probability of the applied probability course - question about meeting within 10 minutes

Problem: Samir and Naomi both arrive at a cafe at a uniformly chosen random time between 9am and 10am. What is the probability that they arrive within ten minutes of each other?

This is a problem of the applied probability course. The explanation used a square to visualize the probabilities, I went at it with a circle and came to 10.9/36 instead of 11/36 and was wondering if I am missing something or my approach is fundamentally flawed.

Visualizing a circle of 60 minutes with 6 equal parts of 10 minutes. Samir has a 1/60th chance of arriving at a given time, if he arrives between 9:10 and 09:50 there is a 2/6th (=10 minutes before and 10 minutes after making 2 * 1/6) chance that Naomie arrives with 10 minutes. So 40 minutes of the hour (40/60) there's a 2/6th chance that they'll meet within 10 minutes.

When Samir arrives at 9:00, there's 1/6th of a chance (10 minutes after). For the times between 9:00-9:10 and 9:50 the time they are able to meet is 10 minutes + 1 minute for every minute after 9:00 and 9:50. So there's 12 minutes they can meet at 9:02 and 9:52 with a chance of 1/60th for each minute. Since this happens at both ends of the hour it's for every 2/60th minute. So this adds up to 10/60+11/60+12/60 .... 17/60+18/60+19/60 = 145/60.

P(09:10-09:50): 40/60 * 20/60 = 800/36.000 P(09:00-09:10 and 09:50-10:00): 145/60 * 2/60 = 290/36.000

800/36.000 + 290/36.000 = 1090/36.000 = 10.9/36.000

Did I miss something, did I make a mistake in my calculations or is there something fundamentally flawed about my reasoning?

I apologize for my less then concise way of explaining my method, I'm just an enthousiast and it takes skill to not have to elaborate!

Note by Max Krakers
1 year, 8 months ago

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Here's how I see it. Assume they can each show up any time within the normalized interval $(0, 1)$. When double-integrating over the interval $(0 \leq x_1 \leq 1)$ and $( 0 \leq x_2 \leq 1 )$ the probability is:

$P = \frac{1}{1^2} \int_0^1 \int_0^1 M \, dx_1 dx_2 \\ M = 1 \,\,\,\, \text{if} \,\, |x_1 - x_2| < \frac{1}{6} \\ M = 0 \,\,\,\, \text{otherwise}$

Numerically approximating this integral results in a probability $P \approx 0.3056$, which matches the expected answer. Your answer is close, but it needs more resolution than just minutes.

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 import math import random N = 10000 dx1 = 1.0/N dx2 = 1.0/N I = 0.0 ###################### x1 = 0.0 while x1 <= 1.0: x2 = 0.0 while x2 <= 1.0: if math.fabs(x1-x2) < 1.0/6.0: I = I + dx1*dx2 x2 = x2 + dx2 x1 = x1 + dx1 ###################### P = I/(1.0**2.0) print N print P ###################### #1000 #0.305277999999 #2000 #0.305861249997 #4000 #0.305569437521 #5000 #0.305677800002 #10000 #0.30556110988 

- 1 year, 8 months ago

The answer is too complex for my understanding, but I understand my approach is an approximation and not a real solution. This is not a problem in the case of the solution the course gives when approaching it as a square of 60 x 60 with a band of 20 minutes acros the square and where the answer is 3600 - the triangles in which they don't meet within 20 minutes.

Thanks!

- 1 year, 8 months ago