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MGMAT Challenge problem of the week

Expert replies
by OneTwoThreeFour » Wed Mar 16, 2011 1:40 pm
The question is from:

https://www.beatthegmat.com/mba/2011/03/ ... 4-mar-2011

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

B

My question:

According to the explanation, the probability of the first point is 1 and the probability of the second point is 1/3, since point 2 can be within 60 degrees from each side of point 1. BUT, in order for point 3 to have 1/6 probability, point 2 must be a point exactly 60 degrees away from either side of point 1. This means that the probability of the second point cannot be 1/3, because 1/3 means that there is a 1/3 chance point 2 can be anywhere within 60 degrees on each side of point 1, when in actuality point 2 can only be a point that is exactly 60 degrees away from either side of the first point. Thus, I believe point 2 cannot have a probability of 1/3.
Its the same thing with the lower limit that has a probability of 1/9. (1 * 1/3 * 1/3) If point 3 can be within 60 degrees away from each side of point 1 (thus the probability of point 3 is 1/3), then point 2 must be on the same point as point 1; this means that point 2 cannot have a probability of 1/3 because it cannot be any point within 60 degrees on each side of point 1, it must be exactly on point 1.

I knows its kind of hard to explain what I am saying without drawing a figure, but can anybody answer my question?
Thanks!
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Source: — Problem Solving |

by kevincanspain » Thu Mar 17, 2011 12:30 am
The difficulty lies in that the probability that the third point lies in the desired range depends where the second point is in relation to the first. We need a simplier approach!
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by sanju09 » Thu Mar 17, 2011 2:48 am
A very convoluted question in the moments of hastening, I rate it as an 800 one. It's been brilliantly explained by Caitlin Clay. Many congratulations.

https://www.beatthegmat.com/mba/2011/03/ ... 4-mar-2011
The mind is everything. What you think you become. -Lord Buddha



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by kevincanspain » Thu Mar 17, 2011 4:07 am
I posted a solution that might be a bit simplier. It's a good thing questions aren't usually so tough on the GMAT!
Kevin Armstrong
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by kevincanspain » Fri Mar 18, 2011 3:23 am
Let's call the points A, B and C. Start at the 12 o'clock position and go clockwise until you find the first point. What is the probability that A is the first of the three points and that the other two are within r units of A? 1/6 x 1/6 = 1/36. Likewise, the probability that B is the first of the three points and that the distance condition is met is 1/36. Finally, the probability that C is the first of the three points and the distance condition is met is 1/36.
These are mutually exclusive events, so we get
1/36 + 1/36 + 1/36 = 1/12
Bear in mind that I assumed that no two points will be at the exact same point on the circle.
Kevin Armstrong
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by OneTwoThreeFour » Fri Mar 18, 2011 9:18 am
Thanks for the alternative explanation Kevin! But could you elaborate on my question? An alternative way to explain it is to lets imagine we have a giant roulette wheel with 360 numbers. The number at the 12 O'Clock position is zero degrees and 360 degrees (Since it is a circle.) and lets assume the numbers increase clockwise. The first ball can land anywhere on the roulette wheel, so it has a probability of 1. Now lets assume it lands on zero. In order for ball 2 to lie no more than a straight line distance away from r, the range that ball 2 can land is from 0 to 60 or 300 to 0. Thus ball 2 will have a probability of 120/360, or 1/3. BUT, in order to have ball 3 land anywhere between 0 and 60 or 300 and 0, ball 2 must land on exactly 60 or 300. Thus, the probability that ball 2 lands on 60 or 360 must be 1/360 + 1/360 = 1/180.

This situation is even more clear for the lower limit of 1/9. Ball 1 can land anywhere on the roulette wheel and lets say it lands again on number 0. In order for ball 3 to have a probability of 1/3, which means that ball 3 can land anywhere from 300 to 60, ball 2 must land exactly on 0 again! If ball 2 does not land on 0, then ball 3 cannot have a probability of 1/3. Thus the probability of ball 2 landing on zero should be 1/360.
Given my reasoning why isn't the answer to the question is a probability from 1/540 <x < 1/1080? (1/540: 1 * 2/360 * 1/3. 1/1080: 1 * 1/360 * 1/3)

Ayudame por favor! If you managed to read it this far, kudos to you.
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by kevincanspain » Fri Mar 18, 2011 11:33 am
OneTwoThreeFour wrote:Thanks for the alternative explanation Kevin! But could you elaborate on my question? An alternative way to explain it is to lets imagine we have a giant roulette wheel with 360 numbers. The number at the 12 O'Clock position is zero degrees and 360 degrees (Since it is a circle.) and lets assume the numbers increase clockwise. The first ball can land anywhere on the roulette wheel, so it has a probability of 1. Now lets assume it lands on zero. In order for ball 2 to lie no more than a straight line distance away from r, the range that ball 2 can land is from 0 to 60 or 300 to 0. Thus ball 2 will have a probability of 120/360, or 1/3. BUT, in order to have ball 3 land anywhere between 0 and 60 or 300 and 0, ball 2 must land on exactly 60 or 300. Thus, the probability that ball 2 lands on 60 or 360 must be 1/360 + 1/360 = 1/180.

You arbitrarily considered degrees: if you had thought that each degree has 60 minutes, you would have calculated a much smaller probability.

Your mistake is analogous to the following flaw in reasoning:

I want to calculate the probability that when two die are rolled, the sum is at least 9. The first dice must show either 1, 2 or 3. The probability that this happens is 1/2. But, in order to have the first dice show anything from 1 to 3, the second must show a 6 (the probability of this is 1/6). Thus the probability of getting a sum of at least 9 is 1/12.


This situation is even more clear for the lower limit of 1/9. Ball 1 can land anywhere on the roulette wheel and lets say it lands again on number 0. In order for ball 3 to have a probability of 1/3, which means that ball 3 can land anywhere from 300 to 60, ball 2 must land exactly on 0 again! If ball 2 does not land on 0, then ball 3 cannot have a probability of 1/3. Thus the probability of ball 2 landing on zero should be 1/360.
Given my reasoning why isn't the answer to the question is a probability from 1/540 <x < 1/1080? (1/540: 1 * 2/360 * 1/3. 1/1080: 1 * 1/360 * 1/3)

Ayudame por favor! If you managed to read it this far, kudos to you.
Suerte con el GMAT! Espero que te haya ayudado
Kevin Armstrong
GMAT Instructor
Gmatclasses
Madrid
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by OneTwoThreeFour » Fri Mar 18, 2011 7:01 pm
Gracias Kevin!

This problem is indeed very tricky since the probability of point 3 happening is both independent and dependent on point 2. We know that point 2 can occur 60 degrees from either side of point 1, but the probability of where point 3 lands depends on where point 2 lands. I then made the mistake of assuming that where point 2 lands is therefore insulated to some fixed points, when in actuality the assumption that point 2 lands exactly 60 degrees away from each side of point 1 and exactly on point 1 is already "included" in the probability of 1/3. In that sense, we can't actually calculate the probability in which all 3 points lie some r distance from one another, but we can definitely estimate that it has to be between 1/9 <x < 1/18. Man, this problem brought me back bad memories of the problem sets I had to do for my Game Theory class.

Anyway, thank you so much once again and viva la Real Madrid!
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by kevincanspain » Sat Mar 19, 2011 8:13 am
You are welcome
Kevin Armstrong
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