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# PS: Light House

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Re: PS: Light House [#permalink]  20 Sep 2010, 22:15
Thanks Bunnel I got it now...Its certainly not a Gmat question.
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Re: PS: Light House [#permalink]  24 Nov 2010, 22:47
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?

I also could not do it in the first go. But here's what I figured out after some thought.

Consider someone standing within the imaginary circle lit by the search-light as it revolves.

What is the probability that the man will stay in dark for 20 seconds (3 revs/min)? zero or (1 - (20/20))
i.e. (1 - P(coming under focus))
What is the probability that the man will stay in dark for 19 seconds? (1 - (19/20))
What is the probability that the man will stay in dark for 18 seconds? (1 - (18/20))
.
.
What is the probability that the man will stay in dark for 5 seconds? (1 - (5/20)) i.e. 3/4

but somehow I am not able to fit in the 'at-least' part. For 'at least' cases, we add the probabilities of all the possible elements (OR). In this case, that would amount to summing up the probabilities of 5 sec, 4 sec, 3 sec, 2 sec and 1 sec.

But I guess the problem is with the discrete approach that I, and many others above have taken.
Shouldn't we approach the problem like the summation of velocity-time graph to find the total distance that we used to do in physics? That is finding the area under the graph..

To be edited, if the bulb in my mind glows, discerning an elegant solution..
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m13#37 [#permalink]  20 May 2011, 21:30
37. 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

can somebody let me know how to solve and reasoning use to find the answer
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Re: m13#37 [#permalink]  20 May 2011, 22:29
Don't know the OA, but...
3 revs in 60 sec -> 1 rev in 20 sec.
5 seconds is one fourth of 20 seconds.
1 revolution is 360 degrees (just to remind: in 20 seconds).
We need one forth, means 360/90=4.
So 1/4 will be an answer.
(A)
What's the OA?
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Re: m13#37 [#permalink]  21 May 2011, 00:13
For sure!
It is said "at least"!
I've found the minimal value.
We need to subtract: 1-1/4=3/4
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Probability - Search Light [#permalink]  06 Sep 2011, 08:38
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. \frac{1}{4}
B. \frac{1}{3}
C. \frac{1}{2}
D. \frac{2}{3}
E. \frac{3}{4}
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Re: Probability - Search Light [#permalink]  06 Sep 2011, 08:56
3 revolutions in 1min=in 60 sec

so 1 rev in 20 sec-> 1/4 rev in 5 sec-> so prob of person remaining in dark for 5 sec is 3/4

so its E
Re: Probability - Search Light   [#permalink] 06 Sep 2011, 08:56
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