BULLS' EYE- Students A, B & C 5

[SidYuca]

This happened in a class. What would you, as a good understanding teacher say to each student, A, B, and C?


Consider two concentric circles with center at O, where the radius of the smaller circle (interior is Black) = 1 and the first drawn larger circle with white interior having radius = 2.

What is the probability that if a point, P, is selected in the larger circle that it is also in the smaller one?

ANSWERS
Student A: "1/4 considering the areas my answer appears correct."

Student B: "1/3 Considering areas my answer appears correct."

Student C: "1/2 Consider drawing the radius OB of the larger circle which passes through point P and intersects the smaller circle at A. Since OA = AB and P is on one of them, my answer appears correct."

 

[SidYuca]

Olaf said

"Do you not consider student C has choosen the radius chord because the radius is the only coordniate deciding if a point is in the inner circle?"

I don't know why C picked the big radius. I do know he picked something that the odds were 50-50 for P distribution on WHAT he picked and gave it as an answer to the problem asked and you and others are supporting him. His error was he didn't use the questions' complete sample space. This is not an error of sampling (it is) but a poor definition. He MAY have been OK taking Inscribed square of each circle(I did not check).

If C said the answer to the problem was 2:1 in favor of the small circle would you agree with his rational... "Draw the diameter AB of the small circle which when extended goes through Point P and Hits at Point C on the Big circle at C? I think not. This too only considers a poor sample of the original population.

Olaf again said
"Any chord you take that will not go through all possible distances to the center from 0 to 2 would not hold as a projection line to any random point and therfore looking at such chords wouldn't give any measurements of the probabillities, would they?"

Two answers:
1. See the segment Described above
2. Correct, No chords wouldn give any measurements of the probabillities because they do not represent the sample space under consideration.

Olaf said:
"I think student C would answer you, he considered a symmetry in this way: Take any random point in the big circle, due to the symmetry of the circle you can always rotate that point to the x-axis for example."

And I'd respond, what does symmetry have to do with it, you could project any to zero on the Z axis?

I want C to be better than his answer. I think he said to his buddies just before the next class after the exam, "Pssst, watch this, I'm going to have some fun with this prof!"

If this figure was a voting district with one party in the big circle and one in the small would you send C out to take your newspapers voting poll or just print the headline, "ELECTION DEAD HEAT

 

[CajunCenturion]

Sid - please draw a picture and label all points accordingly, and post it for everyone. Thank you.

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[CajunCenturion]

==> No really, there are not only poisson, euler, dirac kamm, delta- and several other named and meaningful distributions, you generate and define any distribution and by that rectify all results.
Of course, and is not every one of those distributions mathematically viable? Can you not create a function which produces every one of those distributions? Perhaps we have different understandings of mathematically viable.

==> the one going for a uniform distribution is just the one you would aim for as common sense, if nothing is specified towards the point selection.
You mean the one going for uniform distribution with respect to area is the one you would aim for as common sense. Given that a circle is defined by its radius (All points equidistant from a fixed point) wouldn't common sense dictate that any distribution with circles be based on the ONLY factor (radius) used to define a circle? In other words, a circle is not defined by its area. A circle is defined by its radius and a fixed point.

Given that the problem says absolutely nothing about area, and that a circle is defined solely by its radius and a fixed point, why would one, by default (common sense?) assume that area is the basis for the distribution?

Where does that default assumption of area come from?

==> but the quality of correctness is less good, if you assume ...
The quality of correctness? Is this a fuzzy mathematics problem?
If you assume? Back to assumptions.

==> His error was he didn't use the questions' complete sample space.
Exactly what part of the question's complete sample space has been ignored?

In the original post, student C said this:
"1/2 Consider drawing the radius OB of the larger circle which passes through point P and intersects the smaller circle at A. Since OA = AB and P is on one of them, my answer appears correct."
Again, would you please provide the same picture that student C provided, one with all pertinent features (points, circles, arcs chords, etc) identified. (If you need to, you can e-mail me the graphic and I'll post it)

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[karluk]

As the question is worded, any answer at all between 0% and 100% is correct. I can't tell from the ensuing discussion whether SidYuca would like a do-over and be allowed to restate the problem with more explicit conditions on how the point P is chosen, or whether he feels that the problem is just fine the way it was originally presented. If the latter, then all three students should get full credit, as long as they can show a probability density function that leads to their chosen probabilities. The students can't read the teacher's mind, and I consider it highly inappropriate for the teacher to lecture the students after-the-fact about what they SHOULD have been able to infer. The reality of the matter is that they can't infer anything. If, for example, one makes the plausible assumption this is an approximation to a dart board problem, then the naturual assumption to make is that even a moderately skilled dart thrower should land most of his throws closer to the bull's eye than in the outer regions of the dart board. That would lead to a higher concentration of hits in the inner circle, and B's answer of 1/3 would more likely be right than A's answer of 1/4. With those assumptions, it would be A that deserved criticism for not taking the skill of the player's into consideration in his answer.

 

[SidYuca]

Cajun...
Even if you told me what picture I wouldn't know how to post it. Help?

re. Cajuns' "Given that the problem says absolutely nothing about area, and that a circle is defined solely by its radius and a fixed point, why would one, by default (common sense?) assume that area is the basis for the distribution?"

Yes I appeal to "common sense" because other than defining the circle the radius has zip to do with the circle. On the plane we have three sets, The circle itself, the interior of the circle and the exterior of the circle. I, as most on the planet, appeal to the "common sense" when speaking of problems such as this one. (I, in the future, may have to consider submiting my questions/problems for review first)

I'd never give C credit for his 1:1 answer and justification BUT I might give 100% to a Student D's 1:1 and the following justification had s/he elected to focused ONLY on the 2 circles and NO exteriors/interiors of either when he said:

Student D: "My answer is 1:1. Although the larger circle's circumference is twice the length as the smaller both contain equal degrees of infinity (Aleph)As can be demonstrated by a one to one correspondence "

Karluk
No 'do over' is required or called for by me especially since you now understand my position. For the record I would really like to see your words on how I should have stated the problem.

 

[CajunCenturion]

==> Yes I appeal to "common sense" because other than defining the circle the radius has zip to do with the circle.
Interesting position. How many formulas do you know that are related to circles that do not use the radius?

==> I'd never give C credit for his 1:1 answer and justification BUT I might give 100% to a Student D's 1:1 ...
(I assume that for C you meant 2:1 and that's it's just a typo.)
I'm shocked. You'd give zero credit to an answer that is 100% mathematically correct and provable, yet consider 100% credit to an answer that is completely wrong?

==> Student D: "My answer is 1:1. Although the larger circle's circumference is twice the length as the smaller both contain equal degrees of infinity (Aleph)As can be demonstrated by a one to one correspondence"
First off, the degree of infinity for the number of points is not simply aleph, but aleph-1. The number of points in a circle is a larger than the degree of infinity designated as aleph-0 (aleph-null), but smaller than an even larger infinity known as aleph-2.

Secondly, I want to understand the logic. You would give 100% credit for question on this probability question because both circles have the same number of points?

It's true that both circles have the same number of points. The smaller circle area consists of a set of all points which are a distance <= 1 from the center, and the cardinality (number of members in the set) of that set is indeed aleph-1. The larger circle area consists of a set of points at a distance <= 2 from the center, and the cardinality of that set is also aleph-1. However, even though both sets have the same number of members, the values of those members are not the same. There are values in the larger circle set that are not members of the smaller circle set, and the question is not about the number of members, but about the values of the members. The question is whether a randomly generated value will be a member in both sets (smaller and larger circle set of points) or a member in just the larger circle set of points. Because there are members in one set that do not exist in both sets, any formula that can return at least one value that is only in the larger set of points cannot have a probability of 1:1 of having a value that is in both sets. Why would you give 100% credit for not understanding the difference between the number of members and the values of the members?

Suppose you have the set of even integers and the set of all integers. Both sets have a cardinality of aleph-null (infinite, but demonstrably less than the number of points in our circles), i.e. there are just as many even integers as there are total integers. If the question is what is the probability that a randomly generated integer be a member both sets (set of all integers and the set of even integers) or just a member in the set of all integers? Would you not have to give student D 100% credit for credit for answering 1:1 because both sets contain equal degrees of infinity (aleph-0) as can be demonstrated by a one to one correspondence."

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Back to student's C's response. Is there any point in the larger circle that is being ignored? Is there any point outside of either circle that is being included? You haven't answered my question, "Exactly what part of the question's complete sample space is student C ignoring.?"

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<aside>
I got your e-mail with the diagrams and I'll post them in the morning.
</aside>


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[SidYuca]

CajunCenturion
Yes, a typo, Meant 2:1
It wasn't me that brought the definition of circle in here or how many problems about circles have radii, In the nitty gritty which is so often demanded here the a circle is all the points in the plane a given distance from a given point (called radius and center). It doesn't include the point and it doesn't include the radius. However both are in the interior of the circle.
You are the one who says C is correct not me. So it is not correct for you to say "you give C no points even though he is right."

I did say..."BUT I might give 100% to a Student D's 1:1"
C has shown that any point P in the circle will be in 1/2 or the other and lead to 1:2 ratio on the radius. But that is not the sample space to be examined. I thought that my analogy re. vote polling was excellent (I am biased) But if Republicans lived in large white area and Dems in the smaller black area (NPI) and C went surveying down Radius Lane for voting preferences for the districts and reported 2:1 for the district I'd have a few choice words for him... What the &^%$, you took the bus down Radius lane and asked the passengers, Where did you learn sampling techniques?

Are any points being missed? I don't know. He Demoed point P. He could have elected a different segment also and gotten different results for the segment..i.e diameter of small circle extended through point P continuing to large circle.

 

[CajunCenturion]

Here is the image provided by SidYuca.

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[CajunCenturion]

Since you asked for a response to the voting analogy,
==> I thought that my analogy re. vote polling was excellent
I found the vote polling analogy to be flawed on two counts. First, the smaller circle is a proper subset of the larger circle, that is, every point in the smaller circle is also in the larger circle. The two circles are coincident. That's not the case with the political affiliation of the analogy population. Those two set are disjoint; that is, they have no members in common. The parties are NOT coincident. Secondly, the analogous question would be, "If you voted Republican (defined as being the larger circle), what is the probability that you also voted Democrat (the smaller circle)?" Since both is not a valid answer, the probability is 0%. Therefore, I do not find this to be a valid analogy.

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==> Are any points being missed? I don't know.
If you don't know, then what is the basis for claiming that C isn't using the entire sample space?

==> You are the one who says C is correct not me.
Yes C is correct, and his drawing confirms this.

==> C has shown that any point P in the circle will be in 1/2 or the other and lead to 1:2 ratio on the radius.
What C has shown is that if you draw a straight line from the center point O through the randomly generated point P and continue on the larger circle, then you have a radial segment with a length of the larger circle radius. Half of that radial segment lies in both circles and half lies only in the outer circle. Since P is on that line, there is a 50% chance that it's on the part of the radial that lies in both circles and a 50% chance that is lies on the part that lies only in the outer circle. That is entirely consistent with the drawing submitted by C and is 100% correct.

With respect to the third picture, C didn't use any of those combos, he used a radial that is drawn from the center through point P on out to the outer circle. None of those segments are consistent with C's drawing or description because none of them originate at the center. His description begins with, and I quote from the original, "Consider drawing the radius OB of the larger circle". That means this radial begins at the center O, and continues the edge of the outer circle. None of the segment you show in picture 3 that C "could have" used is a radius OB because none of them start at O. Besides, what does "could have" mean? It's not about what he could have done; it's about what he actually did.

I don't care how you generate point P, you can generate P any way you want. Generate any point P anywhere in the entire sample space. Draw a line from the center of the circle O, that passes through the generated point P and terminates at the outer edge of the larger circle (location B), and label the intersection of that radial with the inner circle as point A. Student C is 100% correct that the length of segment OA equals the length of segment AB. Student C is 100% correct that P has a 50% chance on being on that part of the segment which lies in both circles (OA) and a 50% chance of being on the segment that lies only in the outer circle (AB). Find a point, any point in the entire sample space, where that is not true.

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[CajunCenturion]

Let me try another approach.

Every point in two dimensional space is identified by two components.

For Cartesian coordinates, that point is identified by a X component and a Y component.

With respect to this problem, to identify a point, you randomly determine the x component which ranges from -2 to 2 (+/- the radius of the larger circle), and randomly determine the y component which has the same range of -2 to 2. After you throw out all the points that are outside of the larger circle, you know that all remaining x and y coordinates both lie within the outer circle. The probability that both the x and y coordinates also lie within the inner circle is the probability that x lies within the inner circle (50%) times the probability that y lies within the inner circle (50%) which results in a probability of 25%.

For polar coordinates, that point is identified by a rotational angle [&theta;] and a distance D.

With respect to this problem, to identify a point, you randomly determine a rotation angle [&theta;] which ranges from 0 to 2pi radians (0 to 360 degrees) which encompasses the entire circle, and randomly determine the distance D from the origin which ranges from 0 to 2 (the radius of the larger circle). You don't have to throw out any points since all generated points are already inside the larger circle. The probability that both the rotation angle and distance are also inside the inner circle is the probability that the rotation angle [&theta;] lies within the inner circle (100%) times the probability that the distance D lies within the inner circle (50%) which results in a probability of 50%.

Both solutions are capable of generating any point that lies within either circle using a uniform random number generator, so the full sample space is accounted for. Both answers are completely correct within the coordinate system in which they're calculated.

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[Olaf Doschke]

Yes, Cajun is right.

student C choose the radial line as a good representation of the sample space, so he thinks, as any point P can be seen as a point on some radial line he constructs by: "Consider drawing the radius OB of the larger circle which passes through point P and intersects the smaller circle at A." (see middle of Cajuns image)

However C either wants to "have some fun with this prof!" or just does not see how his reduction of the problem removes an aspect and therefore is no equvalent problem, IF (and only IF) you consider the non specified uniform areal distribution of points.

His choice of choosing P is correct on the other side, as it will always be a point in the bigger circle and half the time in the inner circle. And all that is deterined for point P is, that it is in the larger circle.

Sid, you're right though, that C does not specify he chooses his geometry construction, so it would work for any point P, he considers this as obvious. He also considers it as obvious that any point P will construct either a line OPAB or OAPB, depending on where P is, but indepenadnt of that with OA=AB, as A always is on the redius=1 circle. C does not say he chooses his construction for it's universality for any P.

But C does not see or does see and challenge or bother his professor, that even though his line represents the full sample space of the larger circle, the lengths of oA and AB not necessarily match the probability of points P being either on OA or AB, he rather defines the probability with that.

Confronting student C with other lines, as you did, will not necessarily lead him to rethink this "obvious" assumption the equal line length mean equal probability.

What would perhaps better show student C what he did (if he didn't do it intentionally to bug the professor), is to draw the circle with center O going through P. P can be seen as a representative point for all these points with the same distance to O. Draw this for two different P and you see the amount of points P1 or P2 would represent differs, if the have different distance to O and therefore the probability for their r-coordinate alone might not be equally distributed, at least, if you don't want to define it that way anyway.

Bye, Olaf.

 

[CajunCenturion]

==> But C does not see or does see and challenge or bother his professor, that even though his line represents the full sample space of the larger circle, the lengths of oA and AB not necessarily match the probability of points P being either on OA or AB, he rather defines the probability with that.
I question whether the professor understands polar coordinates. By failing to give proper credit to student C, the professor is punishing the student because the professor lacks the knowledge to understand the answer, or the professor is punishing the student for not assuming unspecified requirements. Either way, the professor is in the wrong.

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[Olaf Doschke]

>Either way, the professor is in the wrong.

Only Sid can say, but I don't think he's denying that the problem specification lacks the determination of a certain distribution of random points and that the professor is in the wrong.

Still student Cs answer lacks pointing out, that he did make use of that lack of specification of a certain point distribution, so he can't prove he has thought of that and is clever or hasn't thought of that and is making an error that only looks clever.

Bye, Olaf.

 

[CajunCenturion]

==> I don't think he's denying that the problem specification lacks the determination of a certain distribution of random points and that the professor is in the wrong.
Maybe, but Sid did say in his post of 27 Nov 11 13:54 that, "I'd never give C credit for his [2]:1 answer and justification".

==> so he can't prove he has thought of that
All you have to do is ask him how he generated his points. If his justification is based on P being generated through polar coordinates, how can you deny him his credit?

It's been clear from several posts that due to a lack of specification on how the points are to be generated, every answer has the potential to be correct, provided that answer is consistent with whatever method was used to generate the points.

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[SidYuca]

I believe you guys are putting me through some kind of pledge hazing! ;>).

Are either of you saying that given 2 circular plates with radius 1:2 that if placed in the rain that one will hold twice as much rain or get twice as wet as the other because the radii are in the ratio of 2:1? Does understanding polar coordinates make this question easier?

This from the source of the problem:
"The 'error' lies in the initial definition of each of two different sample spaces, that is, the set of possible outcomes of an experiment. In the first case,(what I gave as A)^* the sample space is the entire area of the larger circle, while in the second case,(what I gave as C)^* the sample space is the set of points
on OAB. Clearly, when a point is selected on OAB, the probability that the point will be on OA is 1/2. These are two entirely different problems even though (to dramatize the issue) they appear to be the same. The second "solution" is not representative of the problem, that a point is randomly selected from the entire circle. "

^*NOTE: (Not part of the quote)^*

 

[lionelhill]

I'm begining to realise why so few puzzles ever get posted here. The forum ought to be renamed puzzles-for-pedants. Anyone daft enough to stick his/her head above the parapet is likely to be castigated for failing to specify that the puzzle takes place in normal Euclidean surface geometry, and that solutions should exist on the real number line, and not branch off into Quaternions.

Don't expect to post a simple card-trick probability-puzzle without someone refering to Ring Theory and pointing out that the solution doesn't work unless you make the assumption that 1.5 lies between 1.0 and 2.0, which of course it doesn't if you consider distance along the number-line to vary according to the Gehirnquetscher distribution of anticonglomerating numerates.

 

[karluk]

I'm begining to realise why so few puzzles ever get posted here.

That's true. Even though SidYuca has posted two recent problems that turned out to have correct answers that were considerably different than what he anticipated, we should not lose sight of the fact that he single-handedly revived a moribund forum by taking the intiative to post. I will take inspiration by his example and post a problem of my own. I trust I will be able to take it all in good humor, if someone finds a flaw in my logic.

 

[CajunCenturion]

==> Are either of you saying that given 2 circular plates with radius 1:2 that if placed in the rain that one will hold twice as much rain or get twice as wet as the other because the radii are in the ratio of 2:1? Does understanding polar coordinates make this question easier?
No, that's not what I'm saying. What I'm saying is that if you generate a random point using Cartesian coordinates, then the probability of that point being in both circles is 25%. If you generate that point using polar coordinates, then the probability of that point being in both circles is 50%.

The sample space is identical - the set of all points which comprise the larger circle. The difference lies in the algorithm used to generate individual points. A Cartesian coordinate based algorithm will generate a uniform set of points relative to the respective areas of the two circles (4:1). A polar coordinate based algorithm will generate a uniform set of points relative to the respective radii of the two circles (2:1).

==> In the first case,(what I gave as A)^* the sample space is the entire area of the larger circle, while in the second case,(what I gave as C)^* the sample space is the set of points on OAB
As the radial OAB is rotated through the entire 2pi radians (360 degrees) of the circles it will encounter every point that lies within the area of the larger circle; therefore, the two sample spaces are identical.

==> The second "solution" is not representative of the problem, that a point is randomly selected from the entire circle."
Every point in the outer circle lies on some radial OAB. No matter what (x, y) pair you pick from the outer circle, that point can be represented on a OAB radial with an angle [&theta;] = ARCTAN (y/x) and a D = SQRT(x² + y²). Again, this shows that the sample spaces are identical.

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[Olaf Doschke]

lionelhill,

star for the humor alone.

Still it's making fun to see where something drifts.

@Cajun, the polar coordinates are not the culprit, assuming a uniform distribution of radii is.

But again, C didn't say he choose his construction to show, every random point p could be on sme such radial line. Therefore he is reducing the sample space to one single of all these lines. Sid is correct on this. C would not need to change his result, but would need to add to his arguing, that you can do that construcvion for any point P, but he didn't. And again, if C was asked, he would certainly say why he choose such a line to represent tho whole sample space, but he didn't. He made use of many things obvious, but this is like writing program code without any comments on how and why it works.

Bye, Olaf.

 

[SidYuca]

CajunCenturian,

What you are saying is that when you measure the water on the two different moist plates AT THE SAME INSTANT there are at least two different correct answers depending upon how you collect/sample the moisture.