Multiple Sources At Different Frequencies (Probability)

AI Thread Summary
The discussion revolves around calculating the probability that outputs from two sources, A and B, are more recent than each other given their different output frequencies. Source A outputs every 100ms, while source B outputs every 250ms, leading to a potential overlap where A can be ahead of B by up to 50ms. The initial inquiry seeks assistance in determining this probability. After some analysis, it was concluded that the probability of source A being ahead of source B is 0.1. The conversation highlights a desire for a more elegant proof of this finding.
rollypollybear
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Hi

My prob theory is rusty and I am a little embarrassed I can't figure this one out - but:

If we have two sources A and B that are producing an output at different frequencies - say A produces outputs every 100ms and B produces an output every 250ms .. Obviously even if they are perfectly synced, A may be ahead of B at certain times (by up to 50ms)...I am trying to calculate the probability that an event from A will be more recent than B at any point in time

Can someone help me out with some pointers on this?

Cheers
 
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OK I did some scenarios on paper and it looks like the solution is stupidly simple - the probability of source A running at 250ms intervals being ahead of source B running at 100ms intervals is 0.1 ... I don't know if anyone knows how to prove this more elegantly though..
 

Thread 'Formal derivation of statement from Peano Arithmetic system'

I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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