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October 15th, 2011, 01:08 AM  #1 
Newbie Joined: Sep 2011 Posts: 29 Thanks: 0  probability, train stops
A train transports 15 passengers through seven stops. each passenger has equal probability to leave on any of these seven stops and passengers act irrespective of each other  passengers are independent, not couples or married. train stops if someone gets off. What is the probability that no one gets off on first stop? I assume, that probability that someone gets off on some stop is 1/7 and after that 6/7. 1  1/7 = 6/7 And 15 * 6/7 = 12 6/7 = 0.12855 
October 15th, 2011, 10:49 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: probability, train stops
Why do you multiply by 15? That gives a number which clearly isn't a probability (not in the range of 0 to 1). You want to raise to the power of 15.

October 15th, 2011, 11:34 PM  #3 
Newbie Joined: Sep 2011 Posts: 29 Thanks: 0  Re: probability, train stops
Yes, range must be 0 to 1. I assume, that probability that someone gets off on some stop is 1/7 and after that 6/7. 1  1/7 = 6/7 = 0,86 And 0.86^15 / 7 = 0.01487 
October 17th, 2011, 02:38 AM  #4 
Newbie Joined: Sep 2011 Posts: 29 Thanks: 0  Re: probability, train stops
Hello. Is my solution correct?

October 18th, 2011, 12:52 AM  #5 
Newbie Joined: Sep 2011 Posts: 29 Thanks: 0  Re: probability, train stops
I will try again. Can someone answer this, please. I'm just searching a probability that someone gets off on first stop. I assume, that probability that someone gets off on some stop is 1/7 and after that 6/7. 1  1/7 = 6/7 = 0,86 And 0.86^15 / 7 = 0.01487 Maybe this is needed: 1/7 * 2/7 * 3/7 * 4/7 * 5/7 * 6/7 * + 0.01487 ... I don't know, but maybe solution is very near. 
October 19th, 2011, 09:43 AM  #6 
Newbie Joined: Sep 2011 Posts: 29 Thanks: 0  Re: probability, train stops
Hi. Can someone please help. I have really tried to solve and I don't know what a correct solution is. ( I'm afraid that my solution is not correct ).

October 19th, 2011, 03:52 PM  #7  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: probability, train stops Hello, mike688! Quote:
For each passenger:[color=beige] .[/color][color=beige] .[/color] at each stop. We want the probability that all 15 passengers "stay" on the first stop.  
October 20th, 2011, 12:22 AM  #8 
Newbie Joined: Sep 2011 Posts: 29 Thanks: 0  Re: probability, train stops
Thanks a lot! I got confused... Ah, it was just like that. I was close. Now I understand it. Really great that you ( both ) helped.


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