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Title: Bus stop Post by Altamira_64 on Jul 3rd, 2012, 1:58pm A public bus arrives at the bus stop sometime between 6:00 and 6:30. A passenger arrives at the same bus stop some random time within this time interval. What is the probability that the passenger catches the bus within 5 minutes from the time he arrives at the stop? |
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Title: Re: Bus stop Post by towr on Jul 3rd, 2012, 10:23pm I get [hide]11/36, by drawing a diagram[/hide] |
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Title: Re: Bus stop Post by Altamira_64 on Jul 4th, 2012, 1:05am Well, I get 11/72, which one is correct?? |
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Title: Re: Bus stop Post by rmsgrey on Jul 4th, 2012, 3:30am I get 11/72, assuming that the passenger can't catch the bus when the bus arrives first; 11/36 if the bus waits 5 minutes at the stop. |
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Title: Re: Bus stop Post by towr on Jul 4th, 2012, 8:43am I thought it was a bit high... I had (30*30/2-25*25/2)/900, and somehow ended up with 275/900 (or in other words, I neglected to divide 900-625=275 by 2) ::) |
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Title: Re: Bus stop Post by Altamira_64 on Jul 4th, 2012, 9:49am I assume that the bus can't wait for the passenger, if it arrives first. It has to go. So I guess that for the first 25 minutes (25/30 of the total time) the probability is stable, 25/30, that is, 1/6. For the last 5 mins (5/30 of the total time), it decreases linearly and eventually gets to zero. In effect, the total probability is: 25/30 * 1/6 + 5/30 * 1/6 * 1/2 = 11/72. |
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Title: Re: Bus stop Post by rmsgrey on Jul 6th, 2012, 5:10am I solved it with a diagram - plot bus arrival time against passenger arrival time, and you get a strip between tB=tP and tB=tP+5, the area of which is 1/2 - (1/2)(5/6)(5/6) (the triangle from the main diagonal less the triangle from the other diagonal line) or 11/72 |
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Title: Re: Bus stop Post by lopez on Jul 12th, 2012, 3:05am I get 11/72 ::) |
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