challenge question for the week of 03/07/2011-different ans

[kunal227]

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

here is my solution.

lets say the three random, independent points are A, B, C on the circumference of the circle of radius = r

now lets say length of AB = L1
length of BC = L2
length of AC = L3

there are three possibilities L1, L2 and L3 each

each of these lengths can be <r or
=r or
>r

so basically in all there will 27 outcomes, but certain combinations are geometrically impossible and certain others are repititions (because the order of arrangement is not important) so lets see the unique outcomes only

L1, L2, L3
<r, <r, <r ---- desired
=r, =r, =r ---- not possible
>r, >r, >r
<r, <r, =r ---- desired
<r, <r, >r
>r, >r, <r
>r, >r, =r
=r, =r, >r
=r, =r, <r ---- not possible
<r, =r, >r

so a total of 8 unique outcomes possible. out of which only two are the favorable.

there for Probability = 2/8 =1/4.


the answer posted by Manhattan GMAT is still available for your review in the Weekly Challenge Problem section.


Please review and advise. is there any conceptual mistake.

[RonPurewal]

so basically in all there will 27 outcomes, but certain combinations are geometrically impossible and certain others are repititions (because the order of arrangement is not important) so lets see the unique outcomes only

L1, L2, L3
<r, <r, <r ---- desired
=r, =r, =r ---- not possible
>r, >r, >r
<r, <r, =r ---- desired
<r, <r, >r
>r, >r, <r
>r, >r, =r
=r, =r, >r
=r, =r, <r ---- not possible
<r, =r, >r

nope. you're assuming that these events are all equally likely, and that's not true.
in fact, you should realize that this assumption is fatally flawed as soon as you see that some of the possibilities have to be marked "not possible"! i.e., if you use your reasoning, then everything in this list has to have exactly the same probability.

in fact, the probability of any of the events that contain an "=" sign is zero, since those require the placement of points at exactly 60° from each other. in any situation involving probability along a continuum, the probability of any exact location is always 0.
(analogously, if an explosion happens at a completely random time between 3:00 and 4:00 pm, then the probability that the explosion will happen at exactly 3:30:00.00000 pm is zero -- you have to specify a range of times in order to get a nonzero probability.)

if it makes you feel any better, this problem is much, much too hard for the gmat -- in fact, you will notice that the solution key doesn't even present a valid method of exact solution (which, i'm pretty sure, would require calculus). so don't sweat it

[kunal227]

nope. you're assuming that these events are all equally likely, and that's not true.
in fact, you should realize that this assumption is fatally flawed as soon as you see that some of the possibilities have to be marked "not possible"! i.e., if you use your reasoning, then everything in this list has to have exactly the same probability. (yes but i removed those outcomes even from the total number of possible and unique outcomes. Also we if we talk about uniqueness then yes some outcomes can appear more than once in a different order. Even if i consider that then there will be 23 equally likely outcomes out of 27 because 4 of them are not possible to occur geometrically. in that case the probability will become 4/23)

in fact, the probability of any of the events that contain an "=" sign is zero (i dont agree with this because you can have an outcome like this <r,<r,=r ---- you can draw it on the circle and check it), since those require the placement of points at exactly 60° from each other. in any situation involving probability along a continuum, the probability of any exact location is always 0 (only when all three distances are equal to r).
(analogously, if an explosion happens at a completely random time between 3:00 and 4:00 pm, then the probability that the explosion will happen at exactly 3:30:00.00000 pm is zero -- you have to specify a range of times in order to get a nonzero probability.)

if it makes you feel any better (its not about feeling better but getting my concepts straight), this problem is much, much too hard for the gmat (i dont think its too hard, just the solution posted by manhattan makes it look hard) -- in fact, you will notice that the solution key doesn't even present a valid method of exact solution (which, i'm pretty sure, would require calculus). so don't sweat it (i would really like to be proved wrong by an example which wont fit the solution i gave.)

Also the solution posted by manhattan. is just taking the ratio of the location of point on the circle to the entire circle. basically taking ratio of degrees not the outcomes.

[shaji]

The original question and the manhattan solution is flawed.

In mathematics, there is nothing as "approximate probability"
Explaination:
Possible outcomes: 1) The three points are equidistant from any other point by r units.
2) The three points are less than r units from any other point.(favourable outcome)
3)The three points are more than r units from any other point.
Solotion:
The first point has got to be placed and the probabilty is 1
Probailty of placing the second point is 2/6; two favourable outcomes/ total outcomes. Note All three outcomes are equally likely.
Probailty of placing the third point is (2*1)/(2*4); two favourable outcomes/ total outcomes (4 possibilities on either side of the first point.=1/4
Therefore the required exact Probability =1*1/3*1/4-1/12.
The temptation to consider a continuum is the precise trap by the question writer


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so basically in all there will 27 outcomes, but certain combinations are geometrically impossible and certain others are repititions (because the order of arrangement is not important) so lets see the unique outcomes only

L1, L2, L3
<r, <r, <r ---- desired
=r, =r, =r ---- not possible
>r, >r, >r
<r, <r, =r ---- desired
<r, <r, >r
>r, >r, <r
>r, >r, =r
=r, =r, >r
=r, =r, <r ---- not possible
<r, =r, >r

nope. you're assuming that these events are all equally likely, and that's not true.
in fact, you should realize that this assumption is fatally flawed as soon as you see that some of the possibilities have to be marked "not possible"! i.e., if you use your reasoning, then everything in this list has to have exactly the same probability.

in fact, the probability of any of the events that contain an "=" sign is zero, since those require the placement of points at exactly 60° from each other. in any situation involving probability along a continuum, the probability of any exact location is always 0.
(analogously, if an explosion happens at a completely random time between 3:00 and 4:00 pm, then the probability that the explosion will happen at exactly 3:30:00.00000 pm is zero -- you have to specify a range of times in order to get a nonzero probability.)

if it makes you feel any better, this problem is much, much too hard for the gmat -- in fact, you will notice that the solution key doesn't even present a valid method of exact solution (which, i'm pretty sure, would require calculus). so don't sweat it

[kunal227]

Shaji,
Can you provide a more detailed explanation showing ur total outcomes and favorable outcomes?

Can you explain how will you place 3 points on the circumference of a circle with all the three distances equal to radius r? and also with two equal to r and the third less than r? i dont think thats possible. because if any two are equal to r then the third has to greater than r.

[shaji]

Shaji,
Can you provide a more detailed explanation showing ur total outcomes and favorable outcomes?
Please advise you need more explaination for the second or the third point?. Its the third that you need to focus your imagination. I shall be happy to clarify if you need.

Can you explain how will you place 3 points on the circumference of a circle with all the three distances equal to radius r? and also with two equal to r and the third less than r? i dont think thats possible. because if any two are equal to r then the third has to greater than r.

You are correct."if any two are equal to r then the third has to greater than r", in fact third has to be 2*r. Notice that this a unique outcome in itself and does not affect the final probability.

[tim]

Hi folks,
It's really important here to pay attention to what Ron said: first, this is too hard for a GMAT problem; second, any solution that breaks the possibilities down into cases has to ensure that all cases are equally likely. There is nothing wrong with asking for an approximate probability even though there is in fact an exact number out there that is the true probability. But i'm going to give you the real solution and lay this one to rest; any further questions involving cases that are not equally likely will be ignored, as will any questions about my solution that don't demonstrate an understanding of calculus.. :)

First, without loss of generality place the first point at the top of the circle. The second point must be within 60 degrees, which will happen with probability 1/3. Now the probability of the third point being within 60 degrees of both is a function of the distance between the other two, so we integrate the probability with respect to the distance between the first two points. If the first two are 60 degrees apart, there are 60 degrees within which to place the third point. If the first two points coincide, there are 120 degrees within which to place the third point. Because the range of points changes linearly with the distance between the first two, we can take the average of these values to get 90 degrees, which yields a probability of 1/4. So once we hit our initial 1/3 probability, the average probability of the third point sufficing is 1/4, giving us an overall probability of 1/12. i've omitted the rigorous computational details, but this is the general outline..

Again, as Ron promised, the correct answer requires calculus, hence is too difficult to be a real GMAT problem..