## BULLS' EYE- Students A, B & C

## BULLS' EYE- Students A, B & C

(OP)

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."

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."

## RE: BULLS' EYE- Students A, B & C

But I don't claim to be an understanding teacher, I like to make students work

I think I know the correct answer but I am now looking to see if I have a flaw in my reasoning.

## Hidden:

I go with answer A

Computers are like Air conditioners:-

Both stop working when you open Windows

## RE: BULLS' EYE- Students A, B & C

## Spoiler:

This problem is highly reminiscent of the seven digit number puzzle from an earlier thread. SidYucca apparently intended to present that problem with a distribution function that resulted in generating seven digit numbers that were unusually rich in multiples of nine. But he failed to define the problem tightly enough and allowed a smart-aleck student named "karluk" to pick a different distribution that led to some unintended results.

## RE: BULLS' EYE- Students A, B & C

## Hidden:

Student A: correct guess, but you owe a more detailed explaination.

Student B: good thinking, but wrong. The odds are 1:3 and the probabilites have the same relation to each other, but that means 1/4 vs 3/4, not 1/3 vs 2/3, which would translate to odds of 1:2.

Student C: good thinking, but wrong. You're thinking in polar cordinates of distance to O and angle, eleminating the angle as unimportant for the problem, as the inner points merely differ from the outer points in the distance from 0, not in the angle.

But the angle is still important, the polar coordinate system has a spatial uneven nature, with higher distance, same angle delta mean a longer arc, you need to consider that.

Bye, Olaf.

## RE: BULLS' EYE- Students A, B & C

the student would go back, check the work & finally say the could not see anything wrong & ask what they had missed.

the teacher would then say I didn't say there was anything wrong just asked if you were sure about your answer.

Mean but effective.

Computers are like Air conditioners:-

Both stop working when you open Windows

## RE: BULLS' EYE- Students A, B & C

## Hidden:

A - Considers the area of The larger circle with respect to the smaller. 3/4 of the White circle does not contain the black circle. Thus 1/4 of the time point P will be within the Black circle. This would be my preference

B - This one is more difficult for me and might be wrong. Similar to above but considers the shared space differently. 1/4 and 1/4 cancel out? leaving 1/3rd? I can't find a reasonable mathematical reason for this to be true.

C - Is converting a 3 dimentional problem into a 2 dimentional problem. Is ignoring a lot of other space but does not appear to be invalid. Student is defining how he wants to solve the problem and then solving it that way. There could be hundreds of other solutions this way with different results. Example, the circles could be drawn on a sphere introducing other variables not mentioned (or prevented) in the original problem.

I would answer A normally

I would answer C,D,E,F etc if I felt like a smartass

I would not answer B as I cannot fully see the logic.

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

## RE: BULLS' EYE- Students A, B & C

## Hidden:

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

## RE: BULLS' EYE- Students A, B & C

Hoisted/foiled by my own petard! I was thinking of you when I was putting this thing together. The original version had a dart board but I didn't want to explain away missing the dart board, then I thought of putting the circles flat on the bottom of a 4 unit diameter well with no air currents or temerature gradients etc. In my my fervent desire to not hear from you again as in the past and get humbled again you have struck again. Your sharp foil has struck me deep once again. But I am not disappointed by your thrusts I have read all the past puzzles and have read your comments and labeled your reputation. People of your specific talents were always welcome in problem solving meetings I was involved with 20 years ago (before my retirement)to help define and clarify the problem which they always did at any expense. They provided a valuable service at the beginning of each session. Sometimes they continued to contribute to the solution (sometimes not). Sometimes they unilaterally rewrote the problem which did cause some concern.

Please consider solving this problem after inserting my missing "random".

P.S. I think I gave you a star or something but not sure if it went through.

## RE: BULLS' EYE- Students A, B & C

I thought besides the math problem you were asking for a teacher's ability to spot how students are thinking to get to their results and to put their thinking on the right track with the responses to them.

But kudos to karluk, you can of course also bend the problem to match the solution. Not unusual in IT business. ;)

Bye, Olaf.

## RE: BULLS' EYE- Students A, B & C

Bye, Olaf.

## RE: BULLS' EYE- Students A, B & C

"I thought besides the math problem you were asking for a teacher's ability to spot how students are thinking to get to their results and to put their thinking on the right track with the responses to them."

You are exactly correct.

I would also like to see responses to responses.

re. your response to C. I don't know where polar coordinates comes in though. And I don't understand your reference to the angle being a factor in C's thought process. It is the point P (anywhere in the sample space except at O) which defines the radius that Student C references. Note: If P is the center then any radius will serve.

I don't get the thought behind this being a 3D problem.

## RE: BULLS' EYE- Students A, B & C

## Hidden:

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

## RE: BULLS' EYE- Students A, B & C

forum1229: Squaring The Circle

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

## RE: BULLS' EYE- Students A, B & C

## Hidden:

Well, as you say it C is talking of the radius, in polar coordinates that is one part and the other coordinate is the angle.

If you would inverse the problem and say you want to write a program to create random points P you would perhaps start with x,y each in the range of [-2,2] and checking if the point is within the outer and then inner circle, but that makes you loose time for all the points outside. Transform the problem into polar coordinates and the generation of coordinates inside the outer circle is much easier, you create two random numbers radius [0,2] and an angle [0,360[.

That would lead to a distribution of points in the sense of karluk. As the two random numbers are independant and evenly distributed you would get half of those random points inside and half of them outside of the inner circle, because the angle doesn't matter and the evenly distribution of radius would lead to a 1:1 insead of 1:3 distribution of points.

So your problem also has a meaning in generating pseudo random numbers or points, done the wrong way you get a non normal distribution, if you are like most of us, seeing A as the right solution.

The point is: The radius defines the density of points, if you plot a circle with N points, their distance get's higher, the higher the radius is, and therefore a real random distribution of points - as in blindly throwing darts - is not generated by determining random radius and angle, but random x and y.

Write a program determining random points in both ways and see how the distribution of points will differ. It should be significantly visible how uneven the random polar coordinates will distribute generated that way.

Bye, Olaf.

## RE: BULLS' EYE- Students A, B & C

## Hidden:

The angle is indeed eleminated in C's tought process, and that's wrong.

Bye, Olaf.

## RE: BULLS' EYE- Students A, B & C

I am not sure if there is or isn't a 1 to 1 mapping between points in the inner circle and the annulus (I need to study that), but I do know there is a 1 to 1 mapping between the points on the base of a triangle and the segment determined by midpoints on the other two sides (a tease for you) and that the density Question has little or nothing to do with C's response.

I have no doubt that A has the correct answer and fairly sure by his mention "considering the areas" that he 'knows' and no questioning about such things of how, his knowledge of formulas, value/definition of Pi has little to be gained. This is not to say that he didn't cheat!

That being said it appears to me that B needs some further 'talking to/education'. I await the more responses from the group.

As for C, he may be the 'smartest' of the group. His answer is not correct but it is not an uncommon error made by many experimentors and Phd candidates. And I await more comments on his 'experimental' error.

As for Circle Squared, I was a member in it's earliest stages and was 'univited' by its 'leader'. I ocassionally lurk there but not more and that's enough of this.

## RE: BULLS' EYE- Students A, B & C

simply consider generating the random points with radius and angle as in:

## CODE

radius = Random()*2

With Random() being a random number in the range [0..1[ and Random() itself being equal distributed. Then about half of that points have a radius<=1, the angle of course does not matter in respect of inner circle vs annulus.

I made a picture of that and you can see that the distribution is unevenly with a higher weight on inner points than you would consider random:

So while you the coordinates are valid and random, they are not evenly distributed and so it would be "unfair" but would lead to the 1:1 distribution.

You don't need to make the experiment, if you take it for granted random() does generate as many values below .5 than above .5 in the long run. That means the points will be inside and outside the inner circle in about the same amount in the long run.

The distribution looks like this:

http://dl.dropbox.com/u/17129485/points.png

On the left side the points are generated that way, on the right side via random x,y valus, only drawing those points inside the larger circle. I didn't draw the inner r=1 radius, but you get the picture.

Answer C can said to be right in the sense of karluk with that kind of generating the points, as that distribution of points is having a higher densitiy towards the center.

Bye, Olaf.

## RE: BULLS' EYE- Students A, B & C

You missed my point (or was it a hook) when I said:

"I am not sure if there is or isn't a 1 to 1 mapping between points in the inner circle and the annulus (I need to study that), but I do know there is a 1 to 1 mapping between the points on the base of a triangle and the segment determined by midpoints on the other two sides (a tease for you) and that the density Question has little or nothing to do with C's response."

I still am not sure of the first half of my statement above even after seeing your picture and what looked like a darker center of one circle. But darkness should not be confused with lack of unpainted points. I grant you that these 2 methods may produce different 'designs' but back to the point in the problem of one area having more points than the other.

I encourage you to look at my wording of the 'tease' above and tell me your opinion re. my statement of shorter segment vrs longer segment.

## RE: BULLS' EYE- Students A, B & C

## CODE

A uniform distribution is the one where all data points have an equal probability of occurrence. This is the type of distribution created by almost all program psuedo random-number generator functions. Answer "A" presumes a uniform distribution of point selection.

==>

I am not sure if there is or isn't a 1 to 1 mapping between points in the inner circle and the annulusPerhaps we should have a puzzle based on degrees of infinity.

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Good Luck

To get the most from your Tek-Tips experience, please readFAQ181-2886: How can I maximize my chances of getting an answer?

Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

So to make myself clearer I'm talking of a uniform distribution of random values generated by the Random() function. And thus it's easy to see even without studying it in detail, that this kind of random point generation is generating as much points with radius<1 than with <=radius<2 using the formula radius = Random()*2.

And I disagree that "the density Question has little or nothing to do with C's response."

Simple reasoning: C is eleminting the angle from the problem, assuming a uniform distribution of radiusses alone, but a uniform distribution of points in the circle will not lead to such a uniform radius distribution. In fact the distribution of the radiuses of uniformly distributed random points would be radius = Sqrt(Random())*2. Sqrt() being the square root. I teasy you to verify that.

The thought of eleminating the angle due to geometric symmetry is wrong, student C is indirectly introducing a nonuniform distribution with that assumption. For a uniform distribution of points you will need a number of points proportional to the circumference of the radius, which yields morepoints with higher radius than with shorter.

I actually don't understand your other geometry puzzle "there is 1 to 1 mapping between the points on the base of a triangle and the segment determined by midpoints on the other two sides".

In what way do two midpoints determine a segment? At first glance they only define a line. And in what way do you talk of a mapping towards base line points?

Bye, Olaf.

## RE: BULLS' EYE- Students A, B & C

Pardon me, typos like me so much, I can't avoid them.

Bye, Olaf.

## RE: BULLS' EYE- Students A, B & C

Bye, Olaf.

## RE: BULLS' EYE- Students A, B & C

## CODE

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

d, 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.

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Good Luck

To get the most from your Tek-Tips experience, please readFAQ181-2886: How can I maximize my chances of getting an answer?

Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

## RE: BULLS' EYE- Students A, B & C

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?

## RE: BULLS' EYE- Students A, B & C

## Hidden:

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

## RE: BULLS' EYE- Students A, B & C

==>

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.

--------------

Good Luck

To get the most from your Tek-Tips experience, please readFAQ181-2886: How can I maximize my chances of getting an answer?

Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

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.

## RE: BULLS' EYE- Students A, B & C

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.

## RE: BULLS' EYE- Students A, B & C

"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.

## RE: BULLS' EYE- Students A, B & C

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.

Good Luck

To get the most from your Tek-Tips experience, please readFAQ181-2886: How can I maximize my chances of getting an answer?

Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

Degrees = 360 * Rnd

Distance = 2 * Rnd

Good Luck

To get the most from your Tek-Tips experience, please readFAQ181-2886: How can I maximize my chances of getting an answer?

Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

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.

## RE: BULLS' EYE- Students A, B & C

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.

## RE: BULLS' EYE- Students A, B & C

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 θ 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 (θ, d) ] are being uniformly generated.

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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.

Good Luck

To get the most from your Tek-Tips experience, please readFAQ181-2886: How can I maximize my chances of getting an answer?

Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

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.

## RE: BULLS' EYE- Students A, B & C

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.

Good Luck

To get the most from your Tek-Tips experience, please readFAQ181-2886: How can I maximize my chances of getting an answer?

Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

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.

## RE: BULLS' EYE- Students A, B & C

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.

## RE: BULLS' EYE- Students A, B & C

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.

## RE: BULLS' EYE- Students A, B & C

"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"

## RE: BULLS' EYE- Students A, B & C

Good Luck

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Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

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)

Good Luck

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Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

## RE: BULLS' EYE- Students A, B & C

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.

## RE: BULLS' EYE- Students A, B & C

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."

----------

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.?"

----------

<aside>

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

</aside>

Good Luck

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Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

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.

## RE: BULLS' EYE- Students A, B & C

Good Luck

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Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

==>

I thought that my analogy re. vote polling was excellentI 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.

----------

==>

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.

Good Luck

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Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

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 θ and a distance D.

With respect to this problem, to identify a point, you randomly determine a rotation angle θ 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 θ 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.

Good Luck

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Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

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.

## RE: BULLS' EYE- Students A, B & C

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.

Good Luck

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Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

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.

## RE: BULLS' EYE- Students A, B & C

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 thatAll 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.

Good Luck

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Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

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)^*

## RE: BULLS' EYE- Students A, B & C

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.

## RE: BULLS' EYE- Students A, B & C

## RE: BULLS' EYE- Students A, B & C

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 OABAs 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 θ = ARCTAN (y/x) and a D = SQRT(x

^{2}+ y^{2}). Again, this shows that the sample spaces are identical.Good Luck

To get the most from your Tek-Tips experience, please readFAQ181-2886: How can I maximize my chances of getting an answer?

Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

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.

## RE: BULLS' EYE- Students A, B & C

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.

## RE: BULLS' EYE- Students A, B & C

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.This question is about volume and volume requires three dimensions. This is a question about cylinders, not about circles. Further, the original question is about probability distributions which is a completely different exercise.

@Olaf ==> "C didn't say he choose his construction to show, every random point p could be on sme such radial line."

None of the three students explained how the generated their distributions. Why is student C being singled out?

==>

C would not need to change his result,Again, this again shows that you acknowledge that student C's answer is value, since he need not change it. I will grant that you've always acknowledged that his answer was valid.

==>

if C was asked, he would certainly say why he choose such a line to represent tho whole sample space, but he didn't.If C was asked, ... but he didn't. Either C was asked about his methods or he wasn't. How can you possible how he answered a question when the act of asking the question is hypothetical.

It is abundantly clear that default assumptions, whether they be right or wrong, are being used to unfairly discriminate against student C, because his answer, even though correct, is not consistent with those default assumptions. It's also apparent that nothing is going to change that.

Good Luck

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Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

That's why he became the main topic here.

Yes, you could ask everyone to better support their solution and A might then still turn out to guess his correct answer. It would also be interesting to see how B thinks in detail, but since noone here got a grip of what he could have been thinking, he was of less interest here.

I strongly disagree, that I discriminate student C, my hypothetical first answer to student C was "good thinking". But he would need to show if this good thinking I notice was really good thinking, was just cunning, was lazyness, or was perhaps provocation. It's not clear he thought that through.

Bye, Olaf.

## RE: BULLS' EYE- Students A, B & C

Or take a modified version of SidYuca's proposal to measure the collection rate of water on plates of different areas. Let's say that this puzzle is a model for a small, but intense thunderstorm. The rain falls most heavily near the center of the storm, averaging 1" per hour within a mile of the epicenter. But it tapers off thereafter, averaging only 2/3" per hour between one and two miles from the epicenter. Beyond two miles, the rain tapers off to nothing. If the question is asked, "What fraction of the water falls within a mile of the center of the storm, my calculations indicate that the fraction is 1/3, and B's answer becomes the correct one. B has correctly made his calculations based on the relative areas of the region, and has also avoided A's error in not adjusting for the differing intensity of the rain in different parts of the sample space.

Or B may have simply made an error in calculating the relative areas. Which is true? No one knows without asking B, and SidYuca has prevented us from saying definitively that B's answer is wrong by not taking up my offer of a re-do and clarifying how the point P was selected. Instead, he is hoping that everybody else, including the students and us, are clairvoyant enough to figure out that we are required to assume a uniform distribution based on surface area.

Note that SidYuca made almost an identical error in the statement of the "guess the missing digit" puzzle. The only difference was that in that puzzle we were somehow supposed to be clairvoyant enough to assume a HIGHLY non-uniform distribution of the 882 seven digit candidate numbers. So he reserves the right to mark his students wrong, regardless of whether they assume uniform or non-uniform probability distributions.

## RE: BULLS' EYE- Students A, B & C

my hypothetical first answer to student C was "good thinking"Your hypothetical first answer to student C was "good thinking, but wrong".

Good Luck

To get the most from your Tek-Tips experience, please readFAQ181-2886: How can I maximize my chances of getting an answer?

Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

I later admitted karluk is right and C's answer only is wrong assuming a certain distribution of random points. So what?

Still he didn't made that clear so it's doubtful, he actually has thought about really being right with his answer.

Bye, Olaf.

## RE: BULLS' EYE- Students A, B & C

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

## RE: BULLS' EYE- Students A, B & C

1. He IS the person that made two errors in two posts. He is sorry and will try not to repeat in the future.(For this he writes #2.)

2. He wants to know if the following rewrite would have the same type of response(s).

Note: I am not asking for comments re. student responses but you are free to make them.

-------------------re-write is below-----

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 and their interors, 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 a randomly selected point P, in the interior of the larger circle that it is also in interior of 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."

------------------End of rewrite------

## RE: BULLS' EYE- Students A, B & C

By seeing the odds of 1:3 it's clear that probabilities must be a multiples of 1/4, as 1+3=4 parts are the whole area, that comes down to 1/4 vs 3/4 and not 1/3 vs 2/3.

See http://en.wikipedia.org/wiki/Odds on odds vs probabilities.

So even with your Occam's Razor just making that one assumption you would not come out to 1/3 probability.

@Sid, with your rewritten question I would even more stand with my first answers to each student.

Bye, Olaf.

## RE: BULLS' EYE- Students A, B & C

What would you, as a good understanding teacher say to each student, A, B, and C?In all cases I would say the same thing (which is what I have said many times in these fora).

Ok, you've presented your answer, now provide the math to back it it. Make your case and prove your answer.

Either the proof stands up or it doesn't. Everyone one of those answers can be proven mathematically correct on their own merits. I will grant that some will not accept certain answers because they don't conform to some default assumptions or preconceived expectations, but that's not mathematics.

The math will speak for itself and it will either stand up, or it won't. If it does, the student gets full credit, and if it doesn't, then we still down and go through the math and correct any mistakes.

Good Luck

To get the most from your Tek-Tips experience, please readFAQ181-2886: How can I maximize my chances of getting an answer?

Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

I just want an answer key that is correct for a specific question. It is obvious (to me at least) by virtue of my first try that I can make a question that has 3 different answers.

## RE: BULLS' EYE- Students A, B & C

The key point is that if you want a specific answer for a probability question, then you must specify the the domain and its pertinent properties from which the random selection is to be made.

Good Luck

To get the most from your Tek-Tips experience, please readFAQ181-2886: How can I maximize my chances of getting an answer?

Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

With respect to 1/3, it would take me a bit to determine the correct domain and function, but karluk has described the kind of situation with the thunderstorm that you're trying to model.

All that being said, I would consider turning the question around.

Given two concentric circles of radius 1 and 2, provide the mathematics to show that a randomly selected point P that lies in the outer circle has a

Good Luck

To get the most from your Tek-Tips experience, please readFAQ181-2886: How can I maximize my chances of getting an answer?

Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

Bye, Olaf.

## RE: BULLS' EYE- Students A, B & C

The problem with using phrases like "a randomly selected point P" is that "randomly selected" doesn't automatically imply uniform distribution. Consider the implications of using the phrase in the following sentences:

(1) Take a number selected randomly from {2,3,4,5,6,7,8,9,10,11,12} by tossing a pair of dice.

The numbers are selected randomly but, as is well-known, the randomizing device will favor numbers such as 7 over numbers such as 2 or 12.

(2) Take a number selected randomly from {2,3,4,5,6,7,8,9,10,11,12} by tossing a pair of loaded dice.

In case (1), I believe it is reasonable to assume that fair dice are being tossed. Most people would consider it splitting hairs to the nth degree to quibble with the fact that the word "fair" was left out. In case (2), there obviously is no assumption of fairness at all. We would-be problem solvers await further information to more fully determine how the loading of the dice will affect the distribution of the random numbers being generated.

(3) Take a number selected randomly from {2,3,4,5,6,7,8,9,10,11,12} by spinning a wheel divided into 11 equal sized regions and with each region labeled with one of the numbers.

Here we have abandoned dice as our randomizing device and are using a method that will generate the numbers with equal likelihood. Only in case (3) does "selected randomly" equate with "uniform distribution".

## RE: BULLS' EYE- Students A, B & C

if you just specify the geometry of the random experiment but not how it should be done. I made the example of defining a dice with it's 6 faces and asking what the probility of a 6 is. If you don't define you need to throw the dice in such a way you won't have control over the dice, you can also define you simply put the face with the 6 up always.

I already also here suggested "blindly throwing darts", of course only taken all those into account, that hit the board (larger circle area) at all. Sid's rain example with circles on the ground and assumed the rain distribution is uniform is also ok. You can always argue the laws of physics would prefer a certain non normal distribution of darts or would prefer some number on the upper face of a dice after throwing/rolling it, while seeming random, still follows physical laws. Still the core similar thing of executing the random experiment in such ways is the unpredictable outcome.

"Selecting" a point is a bad term in itself, to select suggests to choose with some preferance, not random selection.

Bye, Olaf.

## RE: BULLS' EYE- Students A, B & C

The book then works through two plausible ways to generate a "random chord" and comes up with probabilities of first 1/2 and then 1/3 that the length of the chord is greater than the side of an inscribed triangle.

## RE: BULLS' EYE- Students A, B & C

Your statement:

"Selecting" a point is a bad term in itself, to select suggests to choose with some preferance, not random selection."

Indicates to me that you are beatintg a dead horse and did not read the bold in my post.

further your:

"(1) Take a number selected randomly from {2,3,4,5,6,7,8,9,10,11,12} by tossing a pair of dice.

The numbers are selected randomly but, as is well-known, the randomizing device will favor numbers such as 7 over numbers such as 2 or 12."

I speak of making a random i.e unbiased selection. One can make a random unbiased selection from a bag of 10 red and 4 black checkers. Me thinks you are leaning towards those that complained about the 90% problem.

To all the mathematical purists that gave stories about the interior of the circle when I spoke of circles, you take take more effort in attacking the question rather than its solution. I will in the future will not dwell on defense and that may bring out those that stand on the side watching with great wonderment.

I asked for a rewrite of my question with a single correct answer, why is it that nobody will give that to me?

## RE: BULLS' EYE- Students A, B & C

Professor Ross then poses several questions about this circle, the last of which is

Professor Ross then solves a simple double integral and shows that the answer to question 3 is a^2/R^2. This answers your question, since Professor Ross's Example 1d is the same as your circle problem with a=1 and R=2. Plugging these values into Professor Ross's formula, we get 1^2/2^2 = 1/4 and student A's answer is correct, once the problem has been precisely defined.

## RE: BULLS' EYE- Students A, B & C

I asked for a rewrite of my question with a single correct answer, why is it that nobody will give that to me?Please see my posts of 30 Nov 11 9:37 and 30 Nov 11 9:51.

Good Luck

To get the most from your Tek-Tips experience, please readFAQ181-2886: How can I maximize my chances of getting an answer?

Wise men speak because they have something to say, fools because they have to say something. - Plato## RE: BULLS' EYE- Students A, B & C

Bertrand's original formulation of the paradox involved "randomly selected" chords on a circle. That's undoubtedly the reason for the "random chord" problem I found in my college textbook - Professor Ross was borrowing the example from Bertrand's work. But Olaf and Cajun have shown that the answers of students A and C also represent examples of Bertrand's paradox at work - one gets a different distribution of points depending on whether the "random selection" is on the Cartesian coordinates or polar coordinates of the points in the circle.

## RE: BULLS' EYE- Students A, B & C

An unforeseen consequence of the information revolution has been the exponential propagation of human error.

## RE: BULLS' EYE- Students A, B & C

So if (P) was a dart and you toss it Once it will be in ether to black or white area. So if we limit the dart to a line from centre (0) to A-B then it will be 1:1 it will land in either black or white so "B" is correct for a line

Never give up never give in.

There are no short cuts to anything worth doing