A searchlight on top of the watch-tower makes 3 revolutions

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

1 min=3*360
1 s=18 degree
5 s=90 degree
so probability =1-(90/360)=75%

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.

shouldnt the answer be 3/4?

1-(5/20) atleast 5 includes 5 too. What are the choices?

60 sec => 3 revolutions
5 sec => 1/4 revolution.

Hence 1/4 of the time => Light.
Hence P(darkness) = 3/4.

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?

why is 1/4 of time = light??

also question asks p(darkness for >=5 secs) and not P(darkness)

Bunnel pls explain this...

A searchlight makes 1 revolution in 20 seconds. Consider the diagram below: A man randomly appears at some point M. Now, if beam of light is somewhere in the dark quarter, then the beam will need less than 5 seconds to reach a man and if beam of light is somewhere in the white 3 quarters then it'll need more than 5 seconds to reach a man (so he'll be in the dark more than 5 seconds).

So P=3/4.

Hope it's clear.

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?

We are not assuming that a quadrant is in light. We are assuming that lighted sector has 0 degrees. Anyways not a GMAT question so don't worry about it.

Thanks Bunnel I got it now...Its certainly not a Gmat question.

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..

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?

For sure!
It is said "at least"!
I've found the minimal value.
We need to subtract: \(1-1/4=3/4\)

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