(A) 1/9
(B) 1/12
(C) 1/18
(D) 1/24
(E) 1/27
OA: B
P:S. Not a GMAT question! Pls don't waste time
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GmatKiss
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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
OA: B
P:S. Not a GMAT question! Pls don't waste time
(A) 1/9
(B) 1/12
(C) 1/18
(D) 1/24
(E) 1/27
OA: B
P:S. Not a GMAT question! Pls don't waste time
Last edited by GmatKiss on Sun Oct 16, 2011 5:55 am, edited 1 time in total.
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cans
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the first point can be located anywhere. Now lets call it A.
Now from A, the maximum distance can be 2r and minimum can be 0. Also R distance means 120 degree at centre or 1/3 prob of choosing B.
now if B is the extreme distant point, then 1/6 for C and if B is the nearest point, 1/3 for C
average: 1/3 + 1/6 / 2 = 1/4
thus 1/3*1/4=1/12
Now from A, the maximum distance can be 2r and minimum can be 0. Also R distance means 120 degree at centre or 1/3 prob of choosing B.
now if B is the extreme distant point, then 1/6 for C and if B is the nearest point, 1/3 for C
average: 1/3 + 1/6 / 2 = 1/4
thus 1/3*1/4=1/12
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Cans!!
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Cans!!
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shankar.ashwin
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Honestly it took me a while to even understand the solution. If anyone got this problem on the GMAT, I would consider him or her extremely unlucky.
Consider the 3 points to be A,B and C.
The first point 'A' could be chosen anywhere in the circle, no restrictions.
I am borrowing an image posted by Mitch for a different problem.https://postimage.org/image/11xb37dyc/
Now point B could be chosen anywhere such that the distance from A is between 0 to r
The corresponding probability for the distance (0-r) would be from (0 to 1/3) (0-120 degrees)
Now if point A and B were close to each other (together), point C could similarly be chosen between distance r on either sides or P(1/3)
If B were at a distance 'r' from A, C should be between A and B on the circumference (or between the sector of 60 degree) Prob will be 1/6.
Together we know that the probability will be in-between
1*1/3*1/3 (to) 1*1/3*1/6 (For points A,B and C correspondingly)
(or)
1/9 to 1/18.
Only B satisfies this. I doubt theres even a finite probability for this. I don't think I can explain the solution well enough for anyone to understand, unfortunately.
Consider the 3 points to be A,B and C.
The first point 'A' could be chosen anywhere in the circle, no restrictions.
I am borrowing an image posted by Mitch for a different problem.https://postimage.org/image/11xb37dyc/
Now point B could be chosen anywhere such that the distance from A is between 0 to r
The corresponding probability for the distance (0-r) would be from (0 to 1/3) (0-120 degrees)
Now if point A and B were close to each other (together), point C could similarly be chosen between distance r on either sides or P(1/3)
If B were at a distance 'r' from A, C should be between A and B on the circumference (or between the sector of 60 degree) Prob will be 1/6.
Together we know that the probability will be in-between
1*1/3*1/3 (to) 1*1/3*1/6 (For points A,B and C correspondingly)
(or)
1/9 to 1/18.
Only B satisfies this. I doubt theres even a finite probability for this. I don't think I can explain the solution well enough for anyone to understand, unfortunately.
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GMATGuruNY
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This is NOT a question that would ever appear on the GMAT.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
