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Three points are chosen independently an at random on the circumferenc

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Three points are chosen independently an at random on the circumferenc  [#permalink]

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New post 09 Sep 2015, 00:54
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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

Kudos for a correct solution.

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Three points are chosen independently an at random on the circumferenc  [#permalink]

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New post 09 Sep 2015, 11:05
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Bunuel wrote:
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

Kudos for a correct solution.


Solution: The question means that the angle formed at center with any two of these 3 points should be less than or equal to 60.

First select a point A at random.Then next point B is within 60deg of this point. So, it can be anywhere in between 120deg around A(-60 to 60 from A).
So, prob for B = 120/360 = 1/3.

Selecting C is tricky. The maximum angle for selecting C is 120(When A and B coincide) and minimum angle is 60(When B is 60deg away from this point). As B moves away from A to a maximum of 60,the angle for selecting C linearly decreases from 120 to 60 and as a circle is symmetrical, we can take the average of 60 and 120 for selecting C. So, 90.
Prob for selecting C = 90/360 = 1/4

So, prob of selecting all the points = 1 * 1/3 * 1/4 = 1/12

Option, B
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Three points are chosen independently an at random on the circumferenc  [#permalink]

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New post 13 Sep 2015, 09:48
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Bunuel wrote:
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

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

In order to calculate this probability, we have to treat the placing of each of the three points on the circle as three separate events. The probability of each of these events will then be multiplied together.

The first point can be anywhere on the circle, so the probability is \(\frac{2\pi r}{2\pi r}=1\) . This makes sense, because there are no restrictions on the placement of the first point.

Now that the first point has been placed on the circle, we now have to calculate the probability that the second point will be placed on the circle so that it lays no more than one radius away from the first point. Draw a circle, and place the first point anywhere on that circle. The second point can be a straight-line distance of r away from the first point in either direction. Draw two lines, each with length r, that connect to the circle. The second point can be anywhere within the arc created by the three points drawn on the circle. To calculate the probability of the point being within that arc, we need to know the central angle of the arc. Draw lines connecting the points to the center of the circle. We know that each of those lines will also have a length of r. The circle now contains two triangles, each of which is equilateral. The central angle of each triangle is therefore 60, which means the central angle of the entire arc is 120°.

Image

The probability of placing the second point is therefore 120/360 = 1/3.

Calculating the probability of placing the third point is more difficult. In fact, we can’t calculate the exact probability at all! The probability of placing the third point depends on where the second point is in relation to the first point. We need to look at the two extremes of placement: one in which the second point is one radius away from the first point, and one in which the second point lies as close as possible to the first point.

The probability for the second of these two extremes is easy to calculate: it is the same as the probability of placing the second point. If the second point is as close as possible to the first point, then there is still a 1/3 chance that the third point will lay within one radius of the other two.

Now we need to calculate the probability of placing the third point if the second point is one radius away from the first point. If the first two points are one radius away from each other, then the third point MUST be between them. If the first two points are one radius away from each other, then the minor arc created by the two points has a central angle of 60°.

Image

The probability of placing the third point in this case is then 60/360 = 1/6.

We can now calculate the boundaries of the actual probability. In the first case, the probability of placing all 3 points is 1*1/3*1/3 = 1/9. In the second case, it is 1*1/3*1/6 = 1/18. The probability is therefore somewhere in between these two bounds. The only answer choice that works is 1/12.

The correct answer is B.

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_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Three points are chosen independently an at random on the circumferenc  [#permalink]

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