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A bus stop is served by express buses, which run every hour

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A bus stop is served by express buses, which run every hour [#permalink] New post 04 Jan 2005, 18:45
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A bus stop is served by express buses, which run every hour at ten minutes past the hour, and local buses, which run every quarter of an hour, starting at five past the hour. What is the probability that a passenger arriving at the stop at random will have to wait more than six minutes for a bus?

I have the official answer, however, I am not sure how it was calculated.
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 [#permalink] New post 04 Jan 2005, 22:16
This could be totally wrong but I got 9/20 (45%). The way that I did it was this I assigned a random time and continued on for an hour:

1:05 (local bus), 1:10 (xpress bus), 1:20 (local bus), 1:35 (local bus), 1:50 (local bus), 2:05 (local bus)

Between 1:05 and 1:10 no one would have to wait more than 6 minutes for a bus.

Between 1:10-1:20
1:11, 1:12, 1:13

Between 1:20-1:35
1:21, 1:22, 1:23, 1:24, 1:25, 1:26, 1:27, 1:28 (8 possibilities)

Between 1:35-1:50
(8 possibilities)

Between 1:50-2:05
(8 possibilities)

Probability = (Number of times that you can wait over 6 minutes)/(Total time) = (8*3+3)/60 = 27/60 = 9/20 (45%)

I probably absolutely wrong about this and if I am please let me know.

Thank you.
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 [#permalink] New post 05 Jan 2005, 07:37
greenandwise wrote:
This could be totally wrong but I got 9/20 (45%). The way that I did it was this I assigned a random time and continued on for an hour:

1:05 (local bus), 1:10 (xpress bus), 1:20 (local bus), 1:35 (local bus), 1:50 (local bus), 2:05 (local bus)

Between 1:05 and 1:10 no one would have to wait more than 6 minutes for a bus.

Between 1:10-1:20
1:11, 1:12, 1:13

Between 1:20-1:35
1:21, 1:22, 1:23, 1:24, 1:25, 1:26, 1:27, 1:28 (8 possibilities)

Between 1:35-1:50
(8 possibilities)

Between 1:50-2:05
(8 possibilities)

Probability = (Number of times that you can wait over 6 minutes)/(Total time) = (8*3+3)/60 = 27/60 = 9/20 (45%)

I probably absolutely wrong about this and if I am please let me know.

Thank you.


I did it the same way. It was the only way I could think of so I hope its right.
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 [#permalink] New post 05 Jan 2005, 09:23
Bus leaves the station at 0:05[L] 0:10[E] 0:20[L] 0:35[L] 0.50[L].

A person can arrive at any time so sample space is 60.

Let us calculate the time when he arrives at the station and has to wait for more than 6 mins.

between 0:05 and 0:10 no chance.

0:10 to 0:20 -> only if he arrives between 0:10 and 0:14
therefore 4 minutes--favourable 1

0:20 to 0:35 -> only if he arrives between 0:20 and 0:29
therefore 9 minutes -- favourable 2

0:35 to 0:50 -> only if he arrives between 0:35 and 0:44
therefore 9 minutes -- favourable 3

0:50 to 0:05 --> only if he arrives between 0:50 and 0:59
therefore 9 minutes -- favourable 4

4+9+9+9 31 favourable events

therefore probability is 31/60
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 [#permalink] New post 05 Jan 2005, 10:46
A not-so-easy one for me. Greenandwise, Bhimsen you guys are doing well to solve the question. I think 1:14 should be counted in so I admire Bhimsen's answer.

Now after viewing your answer, I am wondering is there any quicker method to solve the problem? I think I cannot get it done within 2 minutes.

:roll:
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 [#permalink] New post 05 Jan 2005, 12:51
Hmmm...the question asks the probability of waiting more than six minutes for a bus, not 6 minutes or more. This mean that the 6 minute data point is not included (1:14 does not come into the equation) in the number of favorable outcomes. I stand by my 45% probability :-D
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 [#permalink] New post 05 Jan 2005, 13:20
For 40 mins of 60, the probability of waiting > 6 mins exist.hence ans is : 40/60=2/3== 66%.
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 [#permalink] New post 05 Jan 2005, 13:34
Where did you get the 40 mins from? Gmattaker could you please provide the OA because there are a lot of answers?
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PS: Probability [#permalink] New post 05 Jan 2005, 18:50
rthothad is correct. The official answer is 31/60. rthothad, please explain how you calculated this result. Thanks.
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 [#permalink] New post 05 Jan 2005, 19:06
gmattaker, my calculations were as same as Bhimsen's, let me know if you need any clarifications.
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PS: Probability [#permalink] New post 05 Jan 2005, 19:18
My mistake. I missed Bhimsen's solution. Thanks everyone.
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I am also coming with 27/60, and not 31/60 [#permalink] New post 06 Jan 2005, 07:32
Hi, I decided to enumerate all the minutes out as in:

:00
:01
:02
:03
:04
:05 (LOCAL)
:06
:07
:08
:09
:10 (EXPRESS)
:11 over 6 min
:12 over 6 min
:13 over 6 min
:14
:15
:16
:17
:18
:19
:20 (LOCAL)
:21 over 6 min
:22 over 6 min
:23 over 6 min
:24 over 6 min
:25 over 6 min
:26 over 6 min
:27 over 6 min
:28 over 6 min
:29
:30
:31
:32
:33
:34
:35 (LOCAL)
:36 over 6 min
:37 over 6 min
:38 over 6 min
:39 over 6 min
:40 over 6 min
:41 over 6 min
:42 over 6 min
:43 over 6 min
:44
:45
:46
:47
:48
:49
:50 (LOCAL)
:51 over 6 min
:52 over 6 min
:53 over 6 min
:54 over 6 min
:55 over 6 min
:56 over 6 min
:57 over 6 min
:58 over 6 min
:59
:00
:01
:02
:03
:04
:05 (LOCAL)

So if we count all of the over 6 min, we get 27 of them and 27/60=9/20.

I made the assumption that if the commuter arrives at the :10 minute, then it takes him 0 time to take the bus that leaves at the :10 minute. The problem did not state whether or not we can make that assumption. Please advise if this process is correct?

Thanks
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 [#permalink] New post 06 Jan 2005, 10:28
cloaked_vessel you are making a basic mistake....
what if person arrives in between minutes say 0:10:01 or 0:10:55 that is 10 minutes 55 seconds.

You have not accounted for this in your explanation. Please go through my earlier explanation and get back if you still have doubt.
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  [#permalink] 06 Jan 2005, 10:28
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A bus stop is served by express buses, which run every hour

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