My Math Forum probability, train stops

 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:
 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.[color=beige] .[/color]Train stops if someone gets off. What is the probability that no one gets off on first stop?

For each passenger:[color=beige] .[/color]$\begin{Bmatrix}P(\text{get off})=&\frac{1}{7} \\ \\ \\ P(\text{stay})=&\frac{6}{7} \end{Bmatrix}=$[color=beige] .[/color] at each stop.

We want the probability that all 15 passengers "stay" on the first stop.

$\text{The probability is: }\:\left(\frac{6}{7}\right)^{15} \;=\;0.099037155 \;\approx\;\frac{1}{10}$

 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.

 Tags probability, stops, train

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post hbonstrom Applied Math 0 November 17th, 2012 07:11 PM Chikis Physics 2 August 16th, 2012 09:27 PM r-soy Physics 11 June 10th, 2012 01:27 PM token22 Advanced Statistics 2 April 26th, 2012 03:28 PM naspek Calculus 1 December 15th, 2009 01:18 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top