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by kaulnikhil » Wed Mar 10, 2010 11:33 am
If a searchlight on top of a 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


[spoiler]what is the problem if I do 1-1/20-2/20-3/20-4/20=1/2[/spoiler]
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by yeahdisk » Wed Mar 10, 2010 12:31 pm
20 second gap between light

If man appears at any time in the first 15 seconds of this gap he will spend at least 5 seconds in dark

IMO answer is [spoiler]15/20 = 3/4 [/spoiler]
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by kstv » Wed Mar 10, 2010 8:43 pm
The time diff at which a fixed circle will be illuminated = 20 secs.
Let us suppose the man steps into this circle the instance the light has passed.
He will not be caught like a hare on carlight for the next 20 secs.
The total possibility of events = 20.
If he steps in after 15 secs the previous light has passed he will be caught within 5 secs.
So if he steps in any time between 0-15 he is safe for the next 5 secs
So the probability of him staying in the dark is 15/20 = 3/4.

@ kaulnikhil Love this post.
The probability of stepping into the light is same, whether in the 1st min or the 19th minute i.e. 1/20. Independent events.
so 1-{5(1/20)} should also be 3/4
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by Stuart@KaplanGMAT » Thu Mar 11, 2010 12:35 am
kaulnikhil wrote:If a searchlight on top of a 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
First, we have to assume that the light is infinitely thin (i.e. the beam has no thickness).

3 revolutions per minute means that it makes 1 revolution in 20 seconds. Let's divide the circle into 20 equally spaced marks and put the man at the top of the circle (doesn't matter where we put him).

If the light is just past him all the way to the 15 mark, he'll remain in the dark for 5 seconds or more. If the light is between the 15 and the 20 mark, then he'll be illuminated before the 5 seconds is up.

So, there's a 15/20 chance he'll stay dark for 5 or more seconds; choose (e), 3/4.
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by immaculatesahai » Fri Dec 16, 2011 1:58 am
too twisted for my taste. I think I will just click and move on, if I see something of this sort.

Took me more than 5 mins just trying to figure out what to do. No point wasting time on such a problem on the exam. Unless ofcourse we get the very same problem...lol.. then just remember the answer !!!
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by ankush123251 » Fri Dec 16, 2011 2:21 am
@IMMACULATESAHAI -There is nothing difficult about the problem if you read it properly..
The search light makes a complete revolution in 20 seconds.
That means irrespective of where the man enters to find probability:
Probability = Number of favorable outcomes/Total number of outcomes.
Now atleast 5 seconds means a total of 15 seconds(20-5) of darkness.
Thus,
P(darkness) = 15/20 =3/4.
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by immaculatesahai » Fri Dec 16, 2011 2:40 am
@ankush:

When I had encountered the problem I did not assume that the light is extremely thin. We need to assume that to be able to solve the problem. I thought that the light would move on sectors, the way it normally does if you watch a light tower.

Anyway, to each his own I guess. Great that you found it easy :)

My quant is little weak, so I would just click and move on :)
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by ankush123251 » Fri Dec 16, 2011 2:51 am
On a brighter note i did not have the vision of thinking that the light would move in sectors....
Thanks for the update...
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by immaculatesahai » Fri Dec 16, 2011 3:13 am
@Ankush,

Without sounding obsessive, I just wanted to make an illustration regarding what I was saying.

While imagining the problem, I was imagining it as shown in the pictures below.
Attachments
LIght house.JPG
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