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16 Mar 2011, 18:28
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Three points are chosen independently an at random on the circumference of a circle with radius r. What is the approximate probability that none of the three points lies more than a straight-line distance of r away from any other of the three points?
(A) 1/9
(B) 1/12
(C) 1/18
(D) 1/24
(E) 1/27
see the attachment for the official solution done by manhattan gmat instructor.....my solution is given below. I am trying to find out which is correct. I feel that the solution i did is more convincing as the manhattan solution is dividing degrees not the number of possible outcomes. But please feel free to comment.
lets say the three random, independent points are A, B, C on the circumference of the circle of radius = r
now lets say length of AB = L1
length of BC = L2
length of AC = L3
there are three possibilities L1, L2 and L3 each
each of these lengths can be r
so basically in all there will 27 outcomes, but certain combinations are geometrically impossible and certain others are repititions (because the order of arrangement is not important) so lets see the unique outcomes only
L1, L2, L3
r, >r, >r
r
>r, >r, r, >r, =r
=r, =r, >r
=r, =r, r
so a total of 8 unique outcomes possible. out of which only two are the favorable.
there for Probability = 2/8 =1/4.
(A) 1/9
(B) 1/12
(C) 1/18
(D) 1/24
(E) 1/27
see the attachment for the official solution done by manhattan gmat instructor.....my solution is given below. I am trying to find out which is correct. I feel that the solution i did is more convincing as the manhattan solution is dividing degrees not the number of possible outcomes. But please feel free to comment.
lets say the three random, independent points are A, B, C on the circumference of the circle of radius = r
now lets say length of AB = L1
length of BC = L2
length of AC = L3
there are three possibilities L1, L2 and L3 each
each of these lengths can be r
so basically in all there will 27 outcomes, but certain combinations are geometrically impossible and certain others are repititions (because the order of arrangement is not important) so lets see the unique outcomes only
L1, L2, L3
r, >r, >r
r
>r, >r, r, >r, =r
=r, =r, >r
=r, =r, r
so a total of 8 unique outcomes possible. out of which only two are the favorable.
there for Probability = 2/8 =1/4.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
16 Mar 2011, 18:59
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I got 1/24. is this correct?
Posted from my mobile device
Posted from my mobile device
17 Mar 2011, 04:54
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I am looking for the correct answer too. So dont wanna comment on ur answer unless you post the entire solution for it.
Manhattan solution is in the attachment.
Posted from my mobile device
Manhattan solution is in the attachment.
Posted from my mobile device
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
