I had my first day of classes towards my D.Sc in Systems Science and Mathematics. I'm taking "Probability and stochasitc processes". Just to convince us that the idea of "probability" is not trivial (and also that this class could very well kick our collective asses), the professor posed this problem. Super bonus points to any one that can give the complete answer (without cheating on Google).
Start with the unit circle. Randomly select a cord, AB. What is the probability that the length of the cord is greater than sqrt(3)?
If you are brave post your answer in the comments. Extra credit for showing your work. I don't think anybody in the class really "got" the answer. Or at least all of it. I got one part of it, probably the easiest part.
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6 comments:
1/3
An equilateral triangle inscribed in a circle has sides of sqrt(3). So, you can then divide the circle into three equal parts, two sections adjoining point A from which any point creates a chord of less than sqrt(3) and another part across from A in which any point creates a chord greater than sqrt(3).
So, 1 of the 3 sections count - 1/3.
Partial credit.
There are (at least?) two more answers that are "correct".
1/2
Circles have rotational symmetry, so we can "simplify" the problem by normalizing the angle of each cord. If we choose a height on the Y axis, and draw a cord parallel to the X axis, only cords at 1/2 of the heights are longer than sqrt(3).
or 0.3817
If we define our cord as a fixed point on the circumference and any point inside the circle, we find that any point in the 1/3 of the pizza slice circle work, plus any points between the point on the side and the pizza slice (.5*sqrt(3)) for a total of about .3817.
or I could choose an initial point between 1 and infinity, cast rays into the circle to cover all possible cords, and find an answer anywhere between 1/2 and 1/3.
Ok - the upper bound is at least .608.
If I select a point at X=infinity and a random point (with equal distribution) within the area of the circle to define my set of cords, then the set of points that define cords that are too short are the top and bottom crust of the pizza. crust=slice-inner triangle=pi/3-(sqrt(3)/4). number of valid cords = pizza-2 crusts = (pi - crust*2). ratio of valid to total cords = (pi - (pi/3-(sqrt(3)/4))*2) /pi = .608998. What other ways can we count cords?
Okay, I've scooted my window over to the edge of the screen so I can't see anybody else's comments. You pose this question like there are a few secret tricks, but here's my first effort:
1: to select a random chord, create two points, A and B, and assign each a random angular direction from the origin between 0 and 360 degrees from the positive x axis. Draw a line between and we have a random chord AB.
2: It appears that if we inscribe an equilateral triangle on the unit circle, each of its sides has a length of 3^.5. Let's inscribe the triangle so that point A is on one of its vertices. Note that this divides the circle into three equal pieces.
3: Since each point has a uniformly random position relative to our reference frame, I hypothesize they have a uniformly random position relative to each other too. If point B lies farther from point A than the other two vertices of the triangle, chord AB is longer than 3^.5 long. Thus, the probability is 1/3.
So what's the trick?
So the trick is that "randomly select a cord" is an imprecise statement?
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