probability, train stops

mike688 · October 15th, 2011, 02:08 AM

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

CRGreathouse · October 15th, 2011, 11:49 AM

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.

mike688 · October 16th, 2011, 12:34 AM

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

mike688 · October 17th, 2011, 03:38 AM

Hello. Is my solution correct?

mike688 · October 18th, 2011, 01:52 AM

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.

mike688 · October 19th, 2011, 10:43 AM

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 ).

soroban · October 19th, 2011, 04:52 PM

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.

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

mike688 · October 20th, 2011, 01:22 AM

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.

October 15th, 2011, 02:08 AM	#1
mike688 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
$mike688 is offline$

October 15th, 2011, 11:49 AM	#2
CRGreathouse 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.
$CRGreathouse is offline$

October 16th, 2011, 12:34 AM	#3
mike688 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
$mike688 is offline$

October 17th, 2011, 03:38 AM	#4
mike688 Newbie Joined: Sep 2011 Posts: 29 Thanks: 0	Re: probability, train stops Hello. Is my solution correct?
$mike688 is offline$

October 18th, 2011, 01:52 AM	#5
mike688 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.
$mike688 is offline$

October 19th, 2011, 10:43 AM	#6
mike688 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 ).
$mike688 is offline$

October 20th, 2011, 01:22 AM	#8
mike688 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.
$mike688 is offline$

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