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A bus stop is served by express buses, which run every hour [#permalink]
04 Jan 2005, 18:45

00:00

A

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D

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Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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.

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.

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 _________________

I am also coming with 27/60, and not 31/60 [#permalink]
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?