[Guest]

We have a circle (like a target), then two random points of it, A and B, are choosen (like two darts thrown). All of them have the same probability.

1.- If we draw a circle centered in A, with radius to B; What is the probability of it was completely inside of the former?

2.- If we draw the second circle taking the starght line from A to B as its diameter; What is the probability of it was completely inside of the former?

[Aaryan]

Do we not need sizes? Or how large the point is in comparison to the area of the circle? Because technically the number of possibilities are infinite.

[Guest]

So we start with a circle radius r centered at coordinates (0,0), and place points A and B within, each randomly given coordiantes ranging from (-r,-r) to (r,r). The distance between A and B, the radius of the new circle, may be calculated as the hypotenuese of the right triangle formed by their x,y co-rdinates. This is the square root of the squares of the differences between their x and y. This new circle is contained within the larger circle for all values in which the distance from A to the nearest point on the circle is greater than or equal to the new radius. To find that point, draw a line from the center of first circle to point A and on to the circle. This line has total length r. The portion of the line contained by point A and the circle center may be found by treating them as points in a right triangle, thus the length of that line is the square root of Ax^2+Ay^2, and the remaining line connecting point A to the circle is r minus that. So then the new circle is within the old circle for any case of r-(Ax^2+Ay^2)^.5 that is greater than ((Ax-Ax)^2+(Ay-Ay)^2)^.5

r that is greater than ((Ax-Bx)^2+(Ay-By)^2)^.5 + (Ax^2+Ay^2)^.5

So the probability is all A and B which satisfy that equation, divided by the total possiblities for A and B from -r to r

[Guest]

0

for 1) Take any point of a circle. If its the center of another circle, then that circle must stray outside the original circle.

2)The best you can do is have points A and B form a segment that is the diameter of the original. Then the 2nd circle traces the original but is not completely inside of it. If B ventures anywhere else along the circle then once again the 2nd circle must stray outside the original

[Guest]

1. Let r = radius of the circle; 2. randomly choose a point "A" inside the circle; 3. choose "B" randomly within the circle, let the distance of B from the centre of the circle be "x" (x <= r); 4. the favourable location of "B" has got to be such that AB <= (r - x). 5. probability of B being located favourably, pB = (r - x)^2/r^2; 6. probability of A being in the band of x to x+dx, pA = 2*pai*x*dx/(pai*r^2); 7. probability that A and B will together satisfy the desired condition, p = pA * pB; 8. integrating pA*pB for x = 0 to r, we get the desired probability as 1/6; 9. THE ANSWER IS 1/6.

[Guest]

Do we not need sizes? Or how large the point is in comparison to the area of the circle? Because technically the number of possibilities are infinite.

The points are points in a math sense. The circle could be radius 1 to simplify.

[Guest]

1. Let r = radius of the circle; 2. randomly choose a point "A" inside the circle; 3. choose "B" randomly within the circle, let the distance of B from the centre of the circle be "x" (x <= r); 4. the favourable location of "B" has got to be such that AB <= (r - x). 5. probability of B being located favourably, pB = (r - x)^2/r^2; 6. probability of A being in the band of x to x+dx, pA = 2*pai*x*dx/(pai*r^2); 7. probability that A and B will together satisfy the desired condition, p = pA * pB; 8. integrating pA*pB for x = 0 to r, we get the desired probability as 1/6; 9. THE ANSWER IS 1/6.

You are right.

Certainly it involves too much maths. I'm not math expert and I had to think hard to solve it.

[Guest]

0

for 1) Take any point of a circle. If its the center of another circle, then that circle must stray outside the original circle.

2)The best you can do is have points A and B form a segment that is the diameter of the original. Then the 2nd circle traces the original but is not completely inside of it. If B ventures anywhere else along the circle then once again the 2nd circle must stray outside the original

The new circle could be of any size in both cases, according with the random distance from A to B.

[Guest]

0

for 1) Take any point of a circle. If its the center of another circle, then that circle must stray outside the original circle.

2)The best you can do is have points A and B form a segment that is the diameter of the original. Then the 2nd circle traces the original but is not completely inside of it. If B ventures anywhere else along the circle then once again the 2nd circle must stray outside the original

The new circle could be of any size in both cases, according with the random distance from A to B.

I don't understand what this means. My answer is still valid.

A point of a circle will be found along the circumference, not the interior. A circle is set of points that are a given distance from a center point ( C). Therefore, in order to answer this question correctly, one must only consider the points A and B that are r units from C.

In that case, I do not see how the answer can be more than 0

Edited May 17, 2011 by maurice

[Guest]

Sorry, I am not english speaker, and I could explain bad. The points are anywhere inside the circle, not only in the circunference.

[k-man]

Assuming the radius of the original circle is 1.

The probability p(A) that the point A is at least distance x from the circumference of the circle is the area of the circle with radius 1-x divided by the area of the original circle - p(A) = (1-x)²

The probability p(B) that the point B is at most distance x away from the point A is the area of a circle with radius x divided by the area of the original circle - p(B) = x².

The probability of both events occuring is p(A)*p(B) = x²*(1-x)². x can range from 0 to 1, so integrating this function from 0 to 1 we get the result of 1/30.

[k-man]

Looked at Mukul Verma's solution and he is right. My answer is wrong. Good job Mukul!

[Guest]

We have a circle (like a target), then two random points of it, A and B, are choosen (like two darts thrown). All of them have the same probability.

1.- If we draw a circle centered in A, with radius to B; What is the probability of it was completely inside of the former?

2.- If we draw the second circle taking the starght line from A to B as its diameter; What is the probability of it was completely inside of the former?

Easier:

Given the point A (in a circle of radius 1), located at a distance of x from the center:

1.- If we draw a circle centered in A, with radius to a random point of the cricle, B; What is the probability (in terms of x) of it was completely inside of the former?

2.- If we draw the second circle taking the straight line from A to a random point of the circle, B, as its diameter; What is the probability (in terms of x) of it was completely inside of the former?

The first question has been solved:

All points of a circle centered in A, of radius 1-x, accomplish the condition, then probability is the ratio of its area and the major of radius 1: (1-x)^2.

[Guest]

0 for both. points on a circle must lie on the curved line making up the circle not its interior.

[Guest]

0 for both. points on a circle must lie on the curved line making up the circle not its interior.

Sorry for my english, but I mean inside circle.