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.

All this adds up to

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!

#Combinatorics

Note by Max Krakers
1 year, 7 months ago

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Comments

Here's how I see it. Assume they can each show up any time within the normalized interval (0,1)(0, 1) . When double-integrating over the interval (0≤x1≤1) (0 \leq x_1 \leq 1) and (0≤x2≤1)( 0 \leq x_2 \leq 1 ) the probability is:

P=112∫01∫01M dx1dx2M=1    if  ∣x1−x2∣<16M=0    otherwise 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≈0.3056P \approx 0.3056 , which matches the expected answer. Your answer is close, but it needs more resolution than just minutes. 

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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

Steven Chase - 1 year, 7 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!

Max Krakers - 1 year, 7 months ago
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