[BonJoviJones]

So in the "basic question" thread below, the conversation veered to when P(X)=1 and P(X)=0 aren't certainties given infinite sets to work with. I'm intrigued and curious.

Here's my informal proof, anybody want to comment?

(I should note that this started as a "I don't get this post", but half-way through I think I figured it out. I'm quite proud [img]/forums/images/icons/wink.gif[/img])

A guy has a box of colored balls. There is one red ball, and the rest are yellow. If N is the number of balls in the bag, and R is picking a red ball:

P(R) = (lim n->inf) 1/N

This means that P(R)=0 _but_ there is still a red ball in there somewhere, and it can still be drawn.

Am I on track so far?

So let's expand to BruceZ's number example, which I'll rephrase as:

Given an infinite precision random number 0 to 1 inclusive, what is P(Rational Number)?

Using the above ball problem as a template, we get:

P(R) = (rational numbers)/(irrational numbers)

(My calc skills are breaking down, but we want the limit as rational numbers approach infinity, and as irrationtals approach infinity _faster_.)

Given that irrationals are uncountably infinite, while rationals are merely countably infinite it should boil down to P(R) = 0, but like the ball problem it's still possible to get a rational number.

How's that?