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A searchlight on top of the watch-tower makes 3 revolutions per minute. What is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds? (A) 1/4 (B) 1/3 (C) 1/2 (D) 2/3 (E) 3/4
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Is it 3/4....I am not very confident abt the solution but I think that is the answer. If this was a real GMAT test I would have picked that. If this is correct then I will explain.
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marcodonzelli wrote: A searchlight on top of the watch-tower makes 3 revolutions per minute. What is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds?
1/4
1/3
1/2
2/3
3/4 Here is my attempt to solve this one: 3 revolutions per minute = 1 revolution every 20 seconds So no matter what anybody appearing at the tower cannot stay in the dark for more than 20 seconds. This will be our total number of possibilities i.e the denominator. P(man in dark for at least 5 seconds) = 1 - P (man in dark for max of 5 seconds) = 1 - 5/20 = 1 - 1/4 = 3/4 or the other way would be: P(man in dark for at least 5 seconds) is like saying he can be in dark for 5,6,7...all the way to 20 seconds because that is the max. In this approach it would be 15/20 seconds = 3/4. Its kind of a weird approach to solve the problem but I could not think of anything better in the 2mins....
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rotating probability [#permalink]
 01 Sep 2008, 18:40
A searchlight on top of the watch-tower makes 3 revolutions per minute. What is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds?
* \frac{1}{4}
* \frac{1}{3}
* \frac{1}{2}
* \frac{2}{3}
* \frac{3}{4}
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Re: rotating probability [#permalink]
01 Sep 2008, 18:52
Initially I solved the problem for staying in dark for atmost 5 sec. I should read the Q better. Shyt
3 rev- 60 sec means 1 rev - 20 sec
That means the man will become visible under the searchlight with in 20 seconds
Probability that he can stay in the dark for max 5 sec is 5/20 =1/4
Probability that he can stay atleast 5 sec in the dark is 1-1/4 =3/4
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Re: rotating probability [#permalink]
01 Sep 2008, 19:14
yes ... 3/4 is the right ans. such problems are stunners initially, but if properly understood are a cakewalk.
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A searchlight on top of the watch-tower makes 3 revolutions per minute. What is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds?
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Re: PS: Light House [#permalink]
25 Feb 2009, 22:45
1 min=3*360 1 s=18 degree 5 s=90 degree so probability =1-(90/360)=75% bigfernhead wrote: A searchlight on top of the watch-tower makes 3 revolutions per minute. What is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds?
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Re: PS: Light House [#permalink]
25 Feb 2009, 23:44
bigfernhead wrote: A searchlight on top of the watch-tower makes 3 revolutions per minute. What is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds? 3 rpm --> each interval of darkness = 60/3 = 20 s P(atleast 5 s) = 1 - P(less than 5 secs) = 1 - P(<= 4 s) = 1 - (1/20+2/20+3/20+4/20) = 1/2 intuitively this answer seems off, but we can do sanity check by thinking that the lower bound for p(atleast 5 s) is 5/20 = 1/4, additional cases will only increase this value.
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Re: PS: Light House [#permalink]
26 Feb 2009, 07:38
This was my answer, but that's not what the OA says... xyz21 wrote: bigfernhead wrote: A searchlight on top of the watch-tower makes 3 revolutions per minute. What is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds? 3 rpm --> each interval of darkness = 60/3 = 20 s P(atleast 5 s) = 1 - P(less than 5 secs) = 1 - P(<= 4 s) = 1 - (1/20+2/20+3/20+4/20) = 1/2 intuitively this answer seems off, but we can do sanity check by thinking that the lower bound for p(atleast 5 s) is 5/20 = 1/4, additional cases will only increase this value. |
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Re: PS: Light House [#permalink]
27 Feb 2009, 13:17
shouldnt the answer be 3/4?
1-(5/20) atleast 5 includes 5 too. What are the choices?
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Re: PS: Light House [#permalink]
27 Feb 2009, 13:29
The answer is 3/4...
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Re: PS: Light House [#permalink]
28 Feb 2009, 10:59
60 sec => 3 revolutions
5 sec => 1/4 revolution.
Hence 1/4 of the time => Light.
Hence P(darkness) = 3/4.
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Re: PS: Light House [#permalink]
02 Mar 2009, 04:39
my answer is also 3/4.
probabilty to stay in light = 5/20 (3 rev per minutes means 20 sec required for 1 rev).
so, probability to stay in dark is 1-5/20. i.e 3/4.
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Re: PS: Light House [#permalink]
02 Mar 2009, 19:41
abhishekik wrote: my answer is also 3/4.
probabilty to stay in light = 5/20 (3 rev per minutes means 20 sec required for 1 rev).
so, probability to stay in dark is 1-5/20. i.e 3/4. I am not sure if I follow your logic.. * Question does not ask probability to stay in dark, it asks prob of staying in dark for atleast 5 seconds * why is probability to stay in light = 5/20?
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Re: PS: Light House [#permalink]
02 Mar 2009, 19:42
ConkergMat wrote: 60 sec => 3 revolutions
5 sec => 1/4 revolution.
Hence 1/4 of the time => Light.
Hence P(darkness) = 3/4. why is 1/4 of time = light?? also question asks p(darkness for >=5 secs) and not P(darkness)
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Re: PS: Light House [#permalink]
20 Sep 2010, 11:49
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Re: PS: Light House [#permalink]
20 Sep 2010, 13:36
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Re: PS: Light House [#permalink]
20 Sep 2010, 19:00
My doubts - We are not told how much area is in light by light house. We are assuming a quadrant is in light. If light house lights which circles at one revolution per 20 secs only lights an area of narrow beam (30 degree), will the probability be same?
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Re: PS: Light House [#permalink]
20 Sep 2010, 21:05
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Re: PS: Light House
[#permalink]
20 Sep 2010, 21:05
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