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February 23rd, 2019, 03:15 AM  #1  
Newbie Joined: Feb 2019 From: Lithuania Posts: 1 Thanks: 0  Probability of two buyers service time
I have a geometric probability question and I don't know how exactly it should be solved and how graph should look like. The question is: Quote:
Thanks for any help you could give me!  
February 23rd, 2019, 06:12 AM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,417 Thanks: 1025 
So a possibility is: buyer1 could take 8 minutes and buyer2 could take 22 minutes : total 30 minutes ? 
February 23rd, 2019, 10:09 AM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,430 Thanks: 1315  ~
We have to assume that the first customer is served to completion before the second customer begins to be served. With that assumption the total serving time is the sum of the two individual serving times. I'll assume also that the serving time for each customer is independent and are Uniform[8,22], random variables. The PDF of the sum is the convolution of the individual distributions. $p(t) = \dfrac{1}{14^2} (14t30),~16 \leq t \leq 44$ $P[23 \leq t \leq 32] = \displaystyle \int_{23}^{32} p(t) ~dt$ $P[23 \leq t \leq 32] = \dfrac{199}{392} \approx 0.51$ I don't know what you mean by "draw the graph" for the problem. 
March 5th, 2019, 12:53 PM  #4 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 639 Thanks: 85 
Assuming everything uses whole amounts of minutes, you can make a graph with columns with the numbers 14 to 44, with the middle numbers most common kind of like the normal distribution bell curve. The height of each column is the probability, and after calculating all 31 probabilities you could a scale for the graph.


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buyers, probability, service, time 
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