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."

 

[Olaf Doschke]

I tease you, not I teasy you.
Pardon me, typos like me so much, I can't avoid them.

Bye, Olaf.

 

[Olaf Doschke]

Oh, and I agree with you that student C doesn't think about the distribution of the random points, that's why he's making the error in the assumption the radiuses of uniformly distributed points would also have a uniform distribution. The complicated distribution in that aspect is what he does not see, he does not see that his assumed symmetry is not a real symmetry and fails on that.

Bye, Olaf.

 

[CajunCenturion]

@Olaf

[COLOR=white white]I think you're 100% correct for both A and C.
Using Cartesian coordinates and selecting a point with a uniform random x and uniform random y, you'll get answer A.  Using polar coordinates and selecting a point with a uniform random [θ] and uniform random [i]d[/i], you'll get answer C.

Without resorting to a Poisson distribution, I can't see a why to get answer B - yet.

Very nice on the polar coordinates - that's worth a star.[/color]

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[lionelhill]

I greatly admire those who take something complicated and make it simple (it's harder than doing the opposite).

 

[SidYuca]

I gave my thoughts on Student A, so will not repeat here.
Student B, need the most help. It appears to me that he, like A has done the circle calculations correctly but apparently did not consider that the intersection of the two circles were shared by both circles. His thoughts, One black inner circle and annular of the white is equal to 3 of them. Perhaps the teacher can probe about another situation where circles do not intersect, a small black circle, a larger white circle (4x) both on a red background and pose the question about rnd probability re, the 2 circles and not considering in any way the reds.
Student C. It is nice to know about the densities of rnd and polar and deserves the star, however I doubt if the differences would account for the specific answer, 1/2. What else? C's error simply involved the error of confusing the sample spaces. In the problem the sample space is clearly the entire area of the larger circle. For C the sample space was the segment OAB and for that his answer of 1/2 is correct. It's akin to giving results of a US presidential election and only sampling Republican districts or the like.

Re. The segments and the 1 to 1 correspondance. I was thinking about Cantors different degrees of Infinity. Certainly OA and OB probability was 1:1 because they are equal length and a 1:1 correspondace can be set up between the 2 segments I mentioned then perhaps the example should contain a Student D with a problem of Wher OA <> AB with D's arguing for a 1:2 answer also because he could set up a 1:1 correspondace between his 2 segments. I don't know how establish one way or the pther if two different circles have the same degree of infinity.(Will have to read some Cantor)

Olaf. I use drop box but not as you did to show the two diagrams. Could you explain to me or give me a reference?

 

[kwbMitel]

With respect to B's Answer.
[hide]The only reasoning for 1/3 answer that I can fathom is that the space within the white circle that is not black is 3 times the size of the space that is black. As I mentioned before, B seems to be ignoring the fact that the Black space shares with the White that it covers. If the White Space was a torus and not overlapping with Black then I would agree with 1/3. Otherwise, as in this case, no.[/hide]

**********************************************
What's most important is that you realise ... There is no spoon.

 

[CajunCenturion]

When using polar coordinates, there is no error in C's answer. Using polar coordinates, the sample space IS the entire area of the larger circle.

==> I doubt if the differences would account for the specific answer, 1/2.
If the distance value d is generated by a uniform random number generator from 0 to 2, then half the values will be <= 1 and half the values will be > 1, which yields the ration of 1/2.

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[Olaf Doschke]

Sid,

just use the public folder of your dropbox. Then right click on the file you want to share and choose "copy public link" from the dropbox context submenu.

That is using the Windows Explorer plugin.

There also are instructions on how to use the public folder within the public folder.

Bye, Olaf.

 

[Olaf Doschke]

Sid,

I like your arguing for the radius segments OA and AB, but even though both are equal length, they both don't define areas.

You can always argue: From the perspective of uniformly distributed points the circle doesn't play a role at all. Any line going somewhere through the circle has the same probability for random points being on it on all it's points within the circle.

I don't find anything to contradict that very obviously and clearly, only from the persepctive of it's area:

Any 1-dimensional geometric element has an infinite number of points on it, still the nature of it is to have 0 area, and only a certain area>0 can have a probability for points within it.

So actually they have a probability of zero, null, nil to have points on a line, even though any concrete random point will be somewhere on such a line .

You only can make limit calculations towards such a line. For example take a sector of the circle and let it's angle shrink against 0. Once you accept answer A and the symmetry, then 1/4 of uniformly distribuited points will be in the inner segment of radius 1 and 3/4 of the segments points are on the outside part of it, no matter how small you let the angle be.

So in a calculation of a limit towards the segment with an angle of zero, which would be such a radius line, the same radial distribution of points applies, not a uniform distribution.

It's really that way, if you define random point mean uniformly spread in x and y coordinates points with a higher distance to the center do have a higher probability than ones with lower distance, the probability of a point is dependant of it's distance to the center. And that also is true for points on a radial line, they are not equally probable.

Bye, Olaf.

 

[SidYuca]

Cajun,
"When using polar coordinates, there is no error in C's answer."

The problem with that statement is that C's answer to posed question is WRONG. Just as wrong as if the small circle was magnetic and the large (underneah was plastic) and C chose his random numbers by tossing small iron particles. He tried to justify an answer to the original question by focusing the Location of Point P to a line sement and not the original sample space.

OLAF said
"I like your arguing for the radius segments OA and AB, but even though both are equal length, they both don't define areas."

And this exactly what I said. He changed the sample place and he needs to be told that and though what he did was not false it did not apply to the original question.

 

[CajunCenturion]

==> The problem with that statement is that C's answer to posed question is WRONG.
I beg to differ. The question is not asking what is the ratio of the area of the larger circle to the ratio of the smaller circle. It's asking, What is the probability that if a point, P, is selected in the larger circle that it is also in the smaller one?"

Every point in the larger circle can be selected as a vector, i.e, an angle and a distance from the center of the circle. Is that not true?
The angle will range from 0 to 360 degrees. and the distance will range from 0 to 2. The distance cannot exceed 2 because that is the radius of the larger circle. So, using a uniform distribution random number generator, randomly choose an angle 0 to 360 degrees and a distance from 0 to 2.
Degrees = Int(360 * Rnd)
Distance = Int(2 * Rnd)
Every point you pick will be inside the larger circle and half will be inside the smaller circle. All of chosen angles will be in both circles, since both circles contain all 360 degrees. Half of chosen distances (those > 1) will be only in the larger circle) and half of your distances (those <= 1) will be in both circles.

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[CajunCenturion]

Oops - slight correction. You should not be using the Int function.
Degrees = 360 * Rnd
Distance = 2 * Rnd


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[Olaf Doschke]

Yes Cajun,

as the question is not defining how the selection should be made, selecting points the way you describe is valid and will lead to points all within the larger circle and half of them in the smaller one. It leads to a nonuniform distribution, but the question does not disallow that, it's not defining it.

You could reason, that a uniform distribution is the usual case, if nothing is specified. And assuming a uniform distribution of selected points should be made, the selection process would then be invalid.

You can go with polar coordinates for their ease to select pints not outside of the large circle, but you will need to pick the radius nonunifromly then, to have auniform distribution of selected points, with Distance = Sqrt(Rnd)*2.

With that Rnd=.25 leads to a distance of 1 and thus 1/4 of points selected that way will have a distance<=1 and the rest will be in the outer part of the line. And furthermore adding a random uniformly distributed angle the points will be uniformly distributed on the circle, the density will stay equal with the distance, as that nonuniform distribution of distances reflects the skewed nature of polar coordinates and rectifies it.

Bye, Olaf.

 

[Olaf Doschke]

Maybe this is a better clearly understandable observation, thinkning in line lengths:

Eleminating the angle from the problem, as only the distance categorizes points to be either in the inner circle or annulus.

So far correct. But any point you take on the radius line will stand for a circle of points with the same distance, and theses circles have different length. That means while there is a geometric symmetry reducing it from a 2d to a 1d problem this way is not taking that into account.

Again, that is not defined in the question, so it is valid, but you would need to specify what point selection process or point distribution you take into account for your probability.

You can define the selection process as you like, you could simply state "the points should be selected to be in the inner circle with proabability p and outside with 1-p" and then can state any probability.

Still I think student C did not take that into account, he simply ignored the non symmetry of eleminating the angle, and that is to be taken as a flaw in his thinking.

Bye, Olaf.

 

[CajunCenturion]

==> You could reason, that a uniform distribution is the usual case, if nothing is specified. And assuming a uniform distribution of selected points should be made, the selection process would then be invalid.
One could debate whether or not such reasoning is required since, as you say, the problem says nothing about it.

Furthermore, one could argue that a uniform distribution of selected points IS being made.
For Cartesian coordinates ...
Is not the x coordinate being generated uniformly from -2 to 2?
Is not the y coordinate being generated uniformly from -2 to 2?
For Polar coordinates ...
Is not [&theta;] being generated uniformly from 0 to 360?
Is not d being generated uniformly from 0 to 2?

Whether you're using Cartesian coordinates or polar coordinates, both components of a point definition [ (x, y) or ([&theta;], d) ] are being uniformly generated.

==> but you will need to pick the radius non-uniformly then, to have a uniform distribution of selected points, with Distance = Sqrt(Rnd)*2.
That's not solving the presented problem; that's re-defining the problem to achieve a desired result.

==> Eliminating the angle from the problem, as only the distance categorizes points to be either in the inner circle or annulus.
You're no more eliminating the angle from the problem that you would be eliminating the x or y coordinate in a Cartesian solution.

==> But any point you take on the radius line will stand for a circle of points with the same distance,
No it doesn't. A point is a point. In Cartesian coordinates, that point is identified with a x coordinate and a y coordinate. In polar coordinates, that same point is identified with an angle and a distance. Further, you can transform from one coordinate system to the other and back because a point is a point.

The reason that answer A results in 1/4 and answer C results in 1/2 is that for A, you generated points across the diameter (-2 to 2) of the circles and for C, you're generating points across the radius (0 to 2) of the circle. That's the difference in the two answers. Nothing is being eliminated. Both answers are correct, it's all about how you're generate the points.

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[Olaf Doschke]

Cajun,

if you generate polar coordinates with d being generated uniformly from 0 to 2 and plot them you will see there is a density higher towards the center, and while the distribution along d is uniform, it is not uniform in the area.

Yes, I do "re-defining the problem to achieve a desired result."

But I also showed the problem lacks a specification that would lead to any solution being correct. Thus the nature of the problem statement is to be flawed and is meant otherwise. If you don't correct that you can rectify anything.

The real flaw in student C statement is, that he doesn't see how his point selection does make it an nonuniform selection of points. His flaw is, he does not see his projection of all points to a radial line is making a shift of the distribution, and the flaw in his thinking is not, that he made that shift, it's allowed from the outset, but the flaw in his thought is that he does not see he makes this shift.

He doesn't talk about a point distribution, so it's also not really clear that he doesn't take this into account, but it's much more likely. You could as karluk said, especially ask the students on the distribution of points, that will show if student C is or isn't aware of what he does, if projecting the problem geometry to a radial line only.

Bye, Olaf.

 

[CajunCenturion]

==> if you generate polar coordinates with d being generated uniformly from 0 to 2 and plot them you will see there is a density higher towards the center, and while the distribution along d is uniform, it is not uniform in the area.
Yes, you do. The numbers are what they are. You want to disregard them because it's not the answer you're looking for? That doesn't make it wrong.

==> But I also showed the problem lacks a specification that would lead to any solution being correct.
Not necessarily. It still requires all solutions to be mathematically viable. If you can present a solution which meets the conditions of the problem, and can provide the mathematics behind that answer, then you have a valid answer.

==> that he doesn't see how his point selection does make it an nonuniform selection of points.
He doesn't have to; that's not what the question asks for. The question doesn't ask for a method of selecting points so that the ratio of outer circle points to inner circle points is 4:1. The problem acts for the ratio after selecting random points. Even if you assume that the points must be selected from a uniform distribution (which I grant is a quite reasonable requirement), you can still get a valid answer of 2:1 in addition to 4:1.

You're saying that answer C is wrong because the point selection is not uniform with respect to the circle areas. I'm saying that answer C is valid because the point selection is uniform with respect to circle radii.

Why is uniform with respect to area right while uniform with respect to radius wrong? Bottom line, they're both right.

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[Olaf Doschke]

==> But I also showed the problem lacks a specification that would lead to any solution being correct.
Not necessarily. It still requires all solutions to be mathematically viable.

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.

>You want to disregard them because it's not the answer you're looking for? That doesn't make it wrong.

I already twice stated that I agree to that and that any solution is ok, 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.

Why is uniform with respect to area right while uniform with respect to radius wrong? Bottom line, they're both right.

Well, because the commons sense of uniform point distribution is specific to the area distribution and density. Otherwise you can take any distribution and then construct a coordinate system to which this will mean uniform distribution.

You're fighting windmills, Cajun, I am actually not arguing against the things you argue I would be stubborn about. I am stating C's answer can be taken as right, but the quality of correctness is less good, if you assume student C didn't actually see what geometric transformation he made to support his thought.

C is a lazy smartass. He thinks he made a good simplification of the problem, but he didn't. Of course student C is just a construct of Sid, but if it would be a real case, I would strongly assume student C is not aware of the non equvalent transformation of his problem, because I think he didn't have a non uniform areal distribution in mind. I think he really is a bit surprised about his result, actually, but he stays with it due to it's simplicity. Giving him the reasons, why this is a non allowed transformation of the problem - considering an areal uniform distribution - will help him understand what's going on.

Yes, I am tending towards the solutions extending the problem with an uniform distribution area wise, Cajun, but I also already really really early agreed on what karluk said, take any specific distribution and you can get any result, it just has to be anywhere between 0 and 1, to be a probability of course.

But the real problem was not about the math, but about what feedback to give the students, and as no student at all came up with any reference towards distributions that was really just an addon to explain how the different results could be achieved.

You can feedback any student, that he didn't spot the flaw in the problem sepcification and you can say a student like karluk would give the best answer in the first place mathematically.

Translated to IT specifications you could take a customers speciication of a problem and always go back to the customer and stating the specification is not good enough and you're in the position to wait for a more exact specification and it's the fault of the customer things don't go forward.

It's not wrong to make assumption where you are free to, and it's a higher quality if you make reasonable assumptions than unusual ones bending the problem to match your solution.

You can also say student A is quite boring with his solution and C is very creative. It depends what is better applicable.

Bye, Olaf.

 

[SidYuca]

I will try one (or perhaps 2) more times.

I speak to student C, and Proponents of C being correct respond.

"C, I grant you that after you generate your point P (not in the center) and subsequently the radius of the radius OB of the large circle containing P and intersecting the small circle at A that point P is either on OA or OB OF THE RADIUS and that probability is 1/2 For the line. The problem concerned the circle and its interiors, not just the radius.
* Did you consider that points that land on OA are also in The BIG circle?
* If you had instead drawn a chord of the big circle that was also a chord of the small circle and contained point P considered
1. a different answer? And consider that answer to also be correct for the problem
2. Perhaps if the chord described was Tangent
to the small circle? (or contained a miniscule portion of the small circle.?

I think that showing a 1:1 distribution on a subset of the sample space has lead you to make an incorrect generalization for the total sample space.

 

[Olaf Doschke]

Sid,

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? The angle doesn't matter in the categorisation of a point. That's why not any chord could hold for a simplification of the problem. The chord being the radis of the big circle is the best candidate to simplify the problem, no other chord will have that nature.

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.

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?

It's a different story that the projection should perhaps also take into account the rising amount of points towards higher d being projected to the same point on the x-axis, as the circumeference is rising linear to the radius of a circle. We've been through that story in detail already and how it would or would not matter because of no specification of the selection of point P.

Bye, Olaf.