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A Probability Question.


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2 minutes ago, studiot said:

I think OP is referring to Bayesian statistics  (conditional probability).

The notation P(A|B) means P(A), given B.

https://en.wikipedia.org/wiki/Conditional_probability

I agree. But OP seemed to understand the numerator and be only concerned/confused/curious, as the case may be, about the denominator.

And the denominator is P(B)=P(B|A1)+P(B|A2)+P(B|A3)+... for every which Ai in the sample space of the A's.

3 minutes ago, Lorentz Jr said:

I was going to say that! 🤬

+1 😋

Feel free to correct my English anytime, by the way. Conscience is every bit as important as consciousness. ;) 

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18 minutes ago, Genady said:

For example, "anything" can be "not-B" or false. Evidently, it can't, but why?

In this context, "anything" means "any other outcome"  , or "whatever truth value A and/or any other statement besides B might have".

Sorry, let's just leave it at outcomes.

"not B" would be excluded because it's self-referential (actually, because it's inconsistent with B), and "false" isn't an outcome.

Edited by Lorentz Jr
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23 minutes ago, Lorentz Jr said:

In this context, "anything" means "any other outcome"  , or "whatever truth value A and/or any other statement besides B might have".

Sorry, let's just leave it at outcomes.

"not B" would be excluded because it's self-referential (actually, because it's inconsistent with B), and "false" isn't an outcome.

Got it, thanks. But you see why it is confusing.

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1 hour ago, Genady said:

you see why it is confusing.

The OP's usage was confusing. What do "(...|nothing)" and "(...|everything)" mean? "... given that there is no outcome" and "... given that all outcomes occur simultaneously"? Then "(...|anything)" would mean "... given that there is an outcome (which by definition there has to be, and which can be any of the possible outcomes)".*

"anything" is often used in negative contexts in place of "something", i.e. "There's not anything you can do". So "P(B | not anything)" might technically be more accurate. It would mean "There isn't any condition in P(B)", i.e. "For any condition x that excludes any one or more possible outcomes, P(B) is not defined as P(B|x)."

The way @joigus used the expression, it meant "any outcome is acceptable", so his denominator should have been "P(B)=P(B|O1)+P(B|O2)+P(B|O3)+...", because A is a condition but P is a sum over (groups of) outcomes. Writing it in terms of conditions Ai is okay as long as they're both comprehensive and mutually exclusive (i.e. every possible outcome is included exactly once).

 

* ... or "samples" or whatever the right word is. What I have in mind by "outcome" is the set of characteristics of the sample.

Edited by Lorentz Jr
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31 minutes ago, Lorentz Jr said:

The way @joigus used the expression, it meant "any outcome is acceptable", so his denominator should have been "P(B)=P(B|O1)+P(B|O2)+P(B|O3)+...", because A is a condition but P is a sum over (groups of) outcomes. Writing it in terms of conditions Ai is okay as long as they're both comprehensive and mutually exclusive (i.e. every possible outcome is included exactly once).

This is an interesting twist. Denominator expansions should make sense and therefore provide a basis to partition the sample space into significant exclusive propositions. Otherwise, one could end up saying things like, eg,

P(outcome of coin flip = heads) = P(outcome of coin flip = heads| angels have wings) + P(outcome of coin flip = heads| angels do not have wings)

Pretty soon we can get confused by the "logical span" of common language. Angels neither have wings, nor haven't, simply because angels do not exist. We simply cannot assign probabilities to any of both.

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  • 1 year later...

That's an interesting probability question you've got there! In this context, P(B) usually represents the probability of event B occurring, which is not necessarily dependent on "nothing" or "everything." It's more about the likelihood of B happening on its own, given the relevant information or conditions.

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