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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 |
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October 15th, 2011, 10:49 AM | #2 |
Global Moderator Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 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.
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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 |
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October 17th, 2011, 02:38 AM | #4 |
Newbie Joined: Sep 2011 Posts: 29 Thanks: 0 | Re: probability, train stops
Hello. Is my solution correct?
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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. |
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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 ).
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October 19th, 2011, 03:52 PM | #7 | |
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407 | Re: probability, train stops Hello, mike688! Quote:
For each passenger:[color=beige] .[/color] We want the probability that all 15 passengers "stay" on the first stop. | |
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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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