Hugi Magazine 32: Say It With Flowers

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Math Quiz Solutions (By Adok/Hugi)

The previous issue of Hugi featured two tasks about stochastics. In fact it was two tasks which I had previously had to solve for my statistics course at the Vienna University of Technology. I'm happy that these tasks have attracted a bit more people than usually, and I'd like to thank everybody who sent me their solutions: iliks, jkm, mados, Optimus, Stephen Stair and T$. Here are the official solutions.

Task 1.

"Pick two random points on the arc of a circle. What's the probability that the straight line that connects these two points is longer than the radius of the circle?"

Most people have arrived at the solution that this probability is 2/3. This is also the official solution. How do you get there? Draw a circle and mark two random points A and B on it. Also mark the center point C of the circle and draw the lines AB, AC, BC. Now take a look at the angle ACB. If this angle equals 60 degrees, the triangle ACB is equilateral, so "the straight line that connects these two points" AB is equal to the radius of the circle (= AC = BC). If the angle ACB is less than 60 degrees, then AB < radius. If the angle ACB is more than 60 degrees, however, then AB > radius. Since the angle is always equal to 0 or greater and never greater than 180 degrees, AB > radius in 2/3 of the cases.

However, this official solution is not the only imaginable one. Let's take a look at iliks' solution:

"Claus, does the answer to #1 in your Hugi 31 quiz equal 0.5?

How I got it:

You say 'we pick random points'. Here you are not very mathematically strict, because, in fact, probability distribution can be different yet still 'random'. But usually when saying about picking random thing, people have uniform distribution in mind. (Strictly, you should have written, 'we pick two random points uniformly distributed across the circle's arc).

Let r be radius of the circle.

Let l be the length of the horde.

We need to find P{l>r}.

l has two limits:

0 <= l <= 2*r

As we agreed in the beginning, let l be uniformly spread. (We pick random points without any preference.)

So, PDF of l is

f(x) = 1/(2*r), x from [0,2*r]

//(1/(b-a), PDF for uniform distribution, b=2r, a=0)

So, P{l>r} = integral from r to 2r from f(x) = 0.5"

As a matter of fact, this problem is closely related to "Bertrand's paradox". If you check out Bertrand's paradox, you'll see that also for this problem, several correct solutions exist, which, however, lead to different results!

For more information on Betrand's paradox follow these links: [1], [2]

Task 2.

"{a, b} be two numbers out of {1, 2, 3, ..., 20}. (Note: a == b is allowed.) What's the probability that {a, b} have no common factor f? (i.e., they have no common factor f with a mod f == 0 and b mod f == 0)"

This was a bit unclear. What was actually meant was: "What's the probability that {a, b} have no common factor f except 1?"

The answer: 255/400. The denominator comes from the fact that there are 20 * 20 = 400 possible pairs {a, b}. The numerator must be determined by counting/reasoning the number of possible b's for each a.

At the moment I'm not planning to continue this section since I'm not going to have much math at university any more. If somebody wants to take over, feel free to contact me!

Until the next issue, Fable Fox has some questions for you...