What is the probability that a randomly drawn chord will be longer than the radius of the circle?

Prove it.

(In reply to You never know.. by DJ)

okay, I will post the solution in installments, as time allows.

To solve any problem (something you don't yet know) you must first assemble relevant information in which you do have confidence.

1. The probability distribution is flat, that is, there is an equal probability of anything happening (chord choice)

2. There are an infinite number of possible chords (continuous function)

3. the range of chord lengths is 0+ (a really tiny number, right next to zero) through D (the diameter of the circle)

The next thing to do is to think about that range. THe question asks for a pivot point of 1/2 D but we have to ask whether or not a gaussian distribution would in fact center on 1/2 D?

To answer that question, draw a radial line on the circle to an arbitrary intersection to the circle.

What is the chord length at 1/2 R?

in general: h=r-sqrt(4r^2-c^2), where:

h=length of perpendicular bisector of chord of length c to circle or radius r

so let's evaluate this eq for h=1/2 r

c=(sqrt3)r

this means that the probability distribution is not normally centered on 1/2 r. It is centered on (sqrt3)r

Now that that led to something other than an easy solution let's look at that equation again and re-arrange it for c=r:

h=(1-(sqrt3/2))

therefore, from the circle center, (sqrt3)/2 radiis is the pivot point. That is the closer to the center you get from this point the more your chord will be greater than r.

it is further from the center than r/2.

 

Since the probability distribution is flat, then the probability of an event and all events beyond that event is proportional to the relative area and therefore the value itself (normalized).

In our case that value is (sqrt3)/2. It is more than 50% because more than 50% or the chords are larger than the radius.

 

I rely on angular symmetry, that is, the probability of the chord's position on a given radius is equal across all radii.I don't yet agree with the squaring of (sqrt3)/2.

 

Posted by Robert on 2004-06-21 16:03:05