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So I've been thinking about geometric probabilites and have a question that i don't know how to answer.

So let's suppose we have $ABCD$ square with side lengths $1$. And we have a point $P$ inside the square. What is the probability that APB$\angle$ is $90°$? I mean if it's bigger or smaller than that that's an easy task. Since it's bigger if $P$ is inside the Thales circle of $AB$ side, and the angle is smaller if it's outside that. It should be on the circle when it's exactly $90°$. The probability therefore should not be zero (I think), but if you go by the usual understanding it's $\cfrac{\text{good area}}{\text{all area}}$. That half a circle has an area of $0$. Also if you go by the another logic of substracting the probability of an acute and an obtuse angle from $1$, then you also get $0$.

I guess in cases like this you define it differently. But I don't know how. If anyone can help me stir into the right direction or provide mi with some reading on the material that would be highly appreciated. Thank you in advance!

(Also sorry English is not my first language, I might have not used the correct terms everywhere, but hope it's understandable!)

math_inquiry
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    I'm not an expert on this, but I believe the probability will be zero, based on measure theory. Zero probability does not mean impossible. – Deepak Dec 14 '21 at 07:03
  • You various line of thoughts are all good. The probability is indeed zero to have an exact right angle constructed like that. There's nothing wrong or weird about it. – Lee David Chung Lin Dec 14 '21 at 08:42
  • Ok, that is what I thought too, but it is still somewhat weird for me. My point of origin was measure theory too. On the other hand I have a follow up question. By saying that 0 probability not meaning impossible. Do you mean by this, that it can be an arbitraly small number due to our set of possibilities has inifinitely many elements? So basically i was puzzled by this because the 0 probability have different meaning in cominbatorial and geometric probability. At leasdt that's how i understand it now. Thank you for the answers, both were really helpful! – math_inquiry Dec 14 '21 at 09:11

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