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!
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
MGMAT Challenge problem of the week
This topic has expert replies
-
OneTwoThreeFour
- Senior | Next Rank: 100 Posts
- Posts: 85
- Joined: Sat Jan 01, 2011 11:57 am
- Thanked: 1 times
-
kevincanspain
- GMAT Instructor
- Posts: 613
- Joined: Thu Mar 22, 2007 6:17 am
- Location: madrid
- Thanked: 171 times
- Followed by:64 members
- GMAT Score:790
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!
Kevin Armstrong
GMAT Instructor
Gmatclasses
Madrid
GMAT Instructor
Gmatclasses
Madrid
-
sanju09
- GMAT Instructor
- Posts: 3650
- Joined: Wed Jan 21, 2009 4:27 am
- Location: India
- Thanked: 267 times
- Followed by:80 members
- GMAT Score:760
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
https://www.beatthegmat.com/mba/2011/03/ ... 4-mar-2011
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
-
kevincanspain
- GMAT Instructor
- Posts: 613
- Joined: Thu Mar 22, 2007 6:17 am
- Location: madrid
- Thanked: 171 times
- Followed by:64 members
- GMAT Score:790
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
GMAT Instructor
Gmatclasses
Madrid
GMAT Instructor
Gmatclasses
Madrid
-
kevincanspain
- GMAT Instructor
- Posts: 613
- Joined: Thu Mar 22, 2007 6:17 am
- Location: madrid
- Thanked: 171 times
- Followed by:64 members
- GMAT Score:790
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.
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
GMAT Instructor
Gmatclasses
Madrid
GMAT Instructor
Gmatclasses
Madrid
-
OneTwoThreeFour
- Senior | Next Rank: 100 Posts
- Posts: 85
- Joined: Sat Jan 01, 2011 11:57 am
- Thanked: 1 times
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.
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.
-
kevincanspain
- GMAT Instructor
- Posts: 613
- Joined: Thu Mar 22, 2007 6:17 am
- Location: madrid
- Thanked: 171 times
- Followed by:64 members
- GMAT Score:790
Suerte con el GMAT! Espero que te haya ayudadoOneTwoThreeFour 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.
Kevin Armstrong
GMAT Instructor
Gmatclasses
Madrid
GMAT Instructor
Gmatclasses
Madrid
-
OneTwoThreeFour
- Senior | Next Rank: 100 Posts
- Posts: 85
- Joined: Sat Jan 01, 2011 11:57 am
- Thanked: 1 times
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!
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!
-
kevincanspain
- GMAT Instructor
- Posts: 613
- Joined: Thu Mar 22, 2007 6:17 am
- Location: madrid
- Thanked: 171 times
- Followed by:64 members
- GMAT Score:790
