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18 Feb 2019, 13:58
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
35% (02:02) correct
65%
(01:55)
wrong
based on 98
sessions
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GMATH practice exercise (Quant Class 19)
Three points are randomly chosen on the circumference of a given circle. What is the probability that the center of the circle lies inside the triangle whose vertices are at the three points?
(A) 1/3
(B) 1/4
(C) 1/5
(D) 2/5
(E) 2/7
Hint: the problem was created based on another given in the link hidden below
Three points are randomly chosen on the circumference of a given circle. What is the probability that the center of the circle lies inside the triangle whose vertices are at the three points?
(A) 1/3
(B) 1/4
(C) 1/5
(D) 2/5
(E) 2/7
Hint: the problem was created based on another given in the link hidden below
https://gmatclub.com/forum/three-points-are-chosen-independently-an-at-random-on-the-circumferenc-205189.html
Ritu17756
Joined: 30 Sep 2021
Last visit: 25 Apr 2023
Posts: 4
Given Kudos: 16
Location: India
Schools: ISB '23 Kelley '24 McCombs '24
11 Oct 2021, 23:06
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My approach was select any 3 points on the circumference and see when did the center lie inside the triangle.
I realized--- for the center to NOT lie INSIDE my triangle --- the 3 points have to be on the same side of a semi-circle.
i.e. -- imagine an equilateral triangle (points nicely spread out -- center is inside the triangle); compare that to an obtuse triangle on the far side o the circle .
Circumference = 2πr ; semi circle = π2.
1st point can be selected anywhere == Probability =1
2nd point and 3rd point have to be in the same semi circle as the 1st point ===Probability = πr/2πr = 1/2 (each)
Therefore total = 1 * 1/2 * 1/2 = 1/4
I realized--- for the center to NOT lie INSIDE my triangle --- the 3 points have to be on the same side of a semi-circle.
i.e. -- imagine an equilateral triangle (points nicely spread out -- center is inside the triangle); compare that to an obtuse triangle on the far side o the circle .
Circumference = 2πr ; semi circle = π2.
1st point can be selected anywhere == Probability =1
2nd point and 3rd point have to be in the same semi circle as the 1st point ===Probability = πr/2πr = 1/2 (each)
Therefore total = 1 * 1/2 * 1/2 = 1/4
General Discussion
19 Feb 2019, 12:29
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fskilnik
We will refer to the link given in the hint: https://gmatclub.com/forum/three-points ... 05189.html
FOCUS: P(center inside the triangle)
We may consider (without loss of generality) that the first point (blue) was chosen at the position shown in the first figure and that the second point (red) was chosen (without loss of generality) in the upper semicircle (red arc) also presented in the first figure.
In the second figure, we show one possibility of the first two points, to illustrate the fact that we will have a favorable scenario if and only if the third point (green) is chosen in the arc in green, obtained by two lines, each one defined by the center and one of the two first points.
Considering exactly the same reasoning presented in the exercise mentioned in the hint, let´s consider the two extremal cases: when the first two points ("almost") coincide and when the first two points are ("almost") diametrically opposites. The corresponding favorable probabilities are 0 and 1/2, respectively, hence (following the same rationale presented there) our FOCUS is their average: (0+1/2)/2 = 1/4.
The correct answer is (B).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
POST-MORTEM: it is possible to prove that the probabilities of having the first and second points coincidental or diametrically opposites are both ZERO (not approximately zero), therefore there is no trouble omitting the words "almost" presented in parentheses. A subtle consequence: the answer obtained is not approximately right, it is exactly right. (Details are out-of-GMAT´s-scope.)
