GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Aug 2019, 22:25 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. ### Request Expert Reply # Three points are chosen independently an at random on the circumferenc

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 57083
Three points are chosen independently an at random on the circumferenc  [#permalink]

### Show Tags

1
26 00:00

Difficulty:   95% (hard)

Question Stats: 25% (02:49) correct 75% (02:30) wrong based on 104 sessions

### HideShow timer Statistics

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.

_________________
##### Most Helpful Expert Reply
Math Expert V
Joined: 02 Sep 2009
Posts: 57083
Three points are chosen independently an at random on the circumferenc  [#permalink]

### Show Tags

2
3
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°. 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°. 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.

Attachment: d-1.gif [ 4.66 KiB | Viewed 7888 times ]

Attachment: d-2.gif [ 3.92 KiB | Viewed 7893 times ]

_________________
##### General Discussion
Manager  Joined: 10 Aug 2015
Posts: 103
Three points are chosen independently an at random on the circumferenc  [#permalink]

### Show Tags

2
1
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
GMATH Teacher P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 937
Three points are chosen independently an at random on the circumferenc  [#permalink]

### Show Tags

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 Very nice problem!

FOCUS : the probability of having all three points chosen (say blue, red and green, in that order) "sufficiently close".

1. The first point (blue) may be considered (without loss of generality) placed at the top of the circle (first figure).

2. The second point (red) must be placed in the arc of the circle shown in red, somewhere between the two nearer vertices of the top vertex of the regular hexagon (first figure).

Reason: a regular hexagon is composed of 6 equilateral triangles, in this case each side with length r (the radius of the circle that circumscribes the hexagon).

Conclusion: there is a 120/360 = 1/3 probability of choosing the second point favorably.

3. What about the third (green) point? Let´s start considering the two extremal scenarios:

> Extremal Case 1 (where the blue and red points coincide): in this case, we have again 1/3 of probability of choosing the third point in a place such that we have all three points "sufficiently close".

> Extremal Case 2 (where the blue and red points are most apart): in this case, we have 60/360 = 1/6 of probability of choosing the third point in a place such that we have all three points "sufficiently close".

> What about an "Intermediate Case"? If we move the red point away from the blue point, from the Extremal Case 1 until the Extremal Case 2, we have ALL (the infinite continuum possibilities) probabilities from 1/3 to 1/6 possible and, by symmetry, equiprobable!

Conclusion: we must agree that the probability of choosing the third point favorably must be the AVERAGE of 1/3 and 1/6, that is, (1/3 + 1/6)/2 = 1/4.

Finally, the answer we are looking for:

? = P(choosing the second point wisely) * P(choosing the third point wisely given that the second one was chosen wisely) = 1/3 * 1/4 = 1/12.

The correct answer is (B).

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net Three points are chosen independently an at random on the circumferenc   [#permalink] 14 Feb 2019, 18:02
Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

#### MBA Resources  