/sci/ - Science & Math » Thread #14397098

>>14397101
>>14397104
But... but P(A ? B) = P(B ? A), therefor: P(A) + P(B) - P(A?B) = P(B) + P(A) - P(B?A), clean up left and right yields: P(A?B) = P(B?A), am i retarded?

Anonymous

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Anonymous Fri 15 Apr 2022 10:46:40 No.14397135 Report

Quoted By: >>14397152 >>14397172 >>14400241

What is this weird schizo post or complex trolling from everyone in the thread??

Anonymous

Anonymous Fri 15 Apr 2022 10:59:45 No.14397152 Report

Quoted By:

>>14397135
Begone, undergrad.

Anonymous

Anonymous Fri 15 Apr 2022 11:10:53 No.14397172 Report

Quoted By: >>14398324

>>14397135
OP isn't asking about set intersection, he's asking about probabilities

Anonymous

Anonymous Fri 15 Apr 2022 11:39:47 No.14397206 Report

Quoted By: >>14397210 >>14397505

>>14397098
What kind of bait is this?
Of course P(A?B) = P(B?A) is true bro, are you retarded?
Delete this thread.

Anonymous

Anonymous Fri 15 Apr 2022 11:42:01 No.14397210 Report

Quoted By:

>>14397206
Wrong

Anonymous

Anonymous Fri 15 Apr 2022 11:44:48 No.14397216 Report

Quoted By:

There are 8 ips here?
I don't get the joke bros.

Anonymous

Anonymous Fri 15 Apr 2022 13:59:59 No.14397480 Report

Quoted By: >>14397500 >>14398348 >>14400201 >>14406803 >>14406811

Intersection is basically saying "and".
The probability that you get in a car crash and you die is much higher than the probability that you die and then get in a car crash.

Anonymous

Anonymous Fri 15 Apr 2022 14:09:34 No.14397500 Report

Quoted By:

>>14397480
>the first victim

Anonymous

Anonymous Fri 15 Apr 2022 14:13:38 No.14397505 Report

Quoted By: >>14397508

>>14397206
Only if you're in an abelian probability space

Anonymous

Anonymous Fri 15 Apr 2022 14:14:54 No.14397508 Report

Quoted By:

>>14397505
This is true.
$A \wedge B = B \wedge A$ is unprovable from the other axioms of boolean algebras.

Anonymous

Anonymous Fri 15 Apr 2022 20:50:24 No.14398324 Report

Quoted By: >>14399284

>>14397172
probabilities are functions on sets

Anonymous

Anonymous Fri 15 Apr 2022 21:01:32 No.14398348 Report

Quoted By: >>14398354

>>14397480
There is no you die then get in a car crash just as there isn't a car crash then you die. The "and" is that you get in a car crash and die.

Anonymous

Anonymous Fri 15 Apr 2022 21:06:49 No.14398354 Report

Quoted By: >>14399314

>>14398348
Have you ever heard of someone dying and then getting in a car crash?

Anonymous

Anonymous Sat 16 Apr 2022 04:27:46 No.14399284 Report

Quoted By: >>14399499

>>14398324
wrong. Probabilities are how likely something is

Anonymous

Anonymous Sat 16 Apr 2022 04:34:57 No.14399299 Report

Quoted By: >>14399315 >>14399326 >>14402723

A?B being equal to B?A would imply the equivalence of the propositions p & q and q & p. However, take
p: I threw up
q: I ate poop

Then it is clear that "I ate poop and threw up" does not convey the same meaning as "I threw up and ate poop"

Anonymous

Anonymous Sat 16 Apr 2022 04:42:56 No.14399314 Report

Quoted By:

>>14398354
https://www.wiltshire999s.co.uk/crash-driver-dies-wheel-car-thamesdown-drive/

Anonymous

Anonymous Sat 16 Apr 2022 04:43:05 No.14399315 Report

Quoted By:

>>14399299
Logically they are equivalent

Anonymous

Anonymous Sat 16 Apr 2022 04:58:58 No.14399326 Report

Quoted By:

>>14399299
That's just an artifact of the English language. In terms of pure logic the statements are equivalent.

Anonymous

Anonymous Sat 16 Apr 2022 07:27:30 No.14399499 Report

Quoted By: >>14400185

>>14399284
"something" is a set

Anonymous

Anonymous Sat 16 Apr 2022 14:15:46 No.14400185 Report

Quoted By: >>14400613

>>14399499
What if "something" is a proper class?

Anonymous

Anonymous Sat 16 Apr 2022 14:23:56 No.14400201 Report

Quoted By: >>14400205

>>14397480
>and then

Anonymous

Anonymous Sat 16 Apr 2022 14:27:34 No.14400205 Report

Quoted By:

>>14400201
The probability that you then get in a car crash and die is still higher than the probability you die and then get in a car crash

Anonymous

Anonymous Sat 16 Apr 2022 14:47:46 No.14400241 Report

Quoted By:

>>14397135
Not with probability. At least if B can affect A.

Anonymous

Anonymous Sat 16 Apr 2022 17:43:05 No.14400613 Report

Quoted By: >>14400786

>>14400185
still a set

Anonymous

Anonymous Sat 16 Apr 2022 18:53:51 No.14400786 Report

Quoted By: >>14400886

>>14400613
What about the probability of all probabilities that don't contain themselves?

Anonymous

Anonymous Sat 16 Apr 2022 19:02:50 No.14400808 Report

Quoted By: >>14400813

whats upside down u

Anonymous

Anonymous Sat 16 Apr 2022 19:04:40 No.14400813 Report

Quoted By:

>>14400808
and there, no one knows
>WELL UHMM ITS UMM MAGIC

Anonymous

Anonymous Sat 16 Apr 2022 19:30:55 No.14400886 Report

Quoted By: >>14401451

>>14400786
Probabilities aren't sets

Anonymous

Anonymous Sun 17 Apr 2022 00:21:16 No.14401451 Report

Quoted By: >>14402665

>>14400886
yes they are, they are a set of ordered pairs

Anonymous

Anonymous Sun 17 Apr 2022 11:18:54 No.14402665 Report

Quoted By:

>>14401451
nigger

Anonymous

Anonymous Sun 17 Apr 2022 11:51:46 No.14402723 Report

Quoted By:

>>14399299
That's because in English it's implied you mean and then, but you haven't explicitly said it

Anonymous

Anonymous Sun 17 Apr 2022 12:14:26 No.14402765 Report

Quoted By: >>14402770

>>14397098
P(A?B) = P(A) * P(B)
P(B?A) = P(B) * P(A)
P(A) * P(B) = P(B) * P(A)
Thus, we can logicaly conclude that P(A?B) is indeed equal to P(B?A).

What the retards in this threads are confused about is P(A|B) (probability of A knowing B happened) and P(B|A) (probability of B knowing A happened) which are indeed not the same thing although both are included in P(A?B).

Anonymous

Anonymous Sun 17 Apr 2022 12:18:36 No.14402770 Report

Quoted By: >>14405377

>>14402765
>she doesn't know about non-commutative probability spaces

Anonymous

Anonymous Mon 18 Apr 2022 06:00:36 No.14405377 Report

Quoted By:

>>14402770
>made up term for le magical fairyland quantum probability space that is useless for actually modeling anything in the real world
no one cares faggot

Anonymous

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Anonymous Mon 18 Apr 2022 18:45:34 No.14406803 Report

Quoted By: >>14406809 >>14406811

>>14397480
Yeah. I don't get how people don't understand this. Literally the definition of intersection

Anonymous

Anonymous Mon 18 Apr 2022 18:50:39 No.14406809 Report

Quoted By:

>>14406803
You're either baiting or retarded.

Anonymous

Anonymous Mon 18 Apr 2022 18:51:02 No.14406811 Report

Quoted By: >>14406821

>>14397480
>>14406803
By this logic shouldn't the two sets be the same? The first case is when you get into a car crash and die and the second case where you die and get into a car crash refers to the same space, but for some reason you assign chronology to it.

Anonymous

Anonymous Mon 18 Apr 2022 18:54:18 No.14406821 Report

Quoted By:

>>14406811
You're also either baiting or retarded.

Anonymous

Anonymous Tue 19 Apr 2022 03:29:15 No.14408251 Report

Quoted By: >>14408356

>>14397115
bro.

A?B = B?A

already implies

P(A?B) = P(B?A)

>am i retarded?
yes

Anonymous

Anonymous Tue 19 Apr 2022 04:04:24 No.14408356 Report

Quoted By: >>14408393

>>14408251
>A?B = B?A
>already implies
>P(A?B) = P(B?A)
?

Anonymous

Anonymous Tue 19 Apr 2022 04:13:12 No.14408393 Report

Quoted By: >>14408425

>>14408356
A?B = B?A are the same set

Anonymous

Anonymous Tue 19 Apr 2022 04:23:12 No.14408425 Report

Quoted By: >>14408441

>>14408393
but why does that imply P(A?B) and P(B?A) are the same set?

Anonymous

Anonymous Tue 19 Apr 2022 04:29:11 No.14408441 Report

Quoted By: >>14408445

>>14408425
Ok, I can't tell if you mean P as in probability, in which case P(S) is not a set but a real number, or if you mean P as in power set, in which case if A?B = B?A = S, P(S) = P(S) by definition. Do you not mean either of these?

Anonymous

Anonymous Tue 19 Apr 2022 04:30:55 No.14408445 Report

Quoted By: >>14408450

>>14408441
Everything is sets

Anonymous

Anonymous Tue 19 Apr 2022 04:31:58 No.14408450 Report

Quoted By: >>14408454

>>14408445
>Everything is sets
what the fuck do you mean.
I asked you plainly, what does P(set) refer to?
Is it P(S) = The power set of S
Or is it P(S) = The probability some random variable takes on a value in S

Which is it? Or is it something else?

Anonymous

Anonymous Tue 19 Apr 2022 04:33:22 No.14408454 Report

Quoted By: >>14408465 >>14408519

>>14408450
the probability of A?B
and everything is sets, real numbers are just sets of dedekin cuts, which are sets of rational numbers, which are ordered pairs (which are sets) of natural numbers, which are sets built up out of empty set

Anonymous

Anonymous Tue 19 Apr 2022 04:37:40 No.14408465 Report

Quoted By: >>14409333

>>14408454
>everything is sets
please show me a proof of the consistency of ZFC lol, but anyways,
A?B = B?A = S by the definition of intersection.
Again by definition, P(A?B) = P(B?A) = P(S).

Its literally that simple. The intersection of A and B is "the set of elements that belong to both the set A and the set B" and the intersection of B and A is "the set of elements that belong to both the set B and A." These are the same set, call that set S. The probability that some random variable takes on a value in S IS EQUAL TO the probability that random variable takes on a value in S.
P(S) = P(S).

I don't understand what the issue is. Are you thinking of the probability of an event A GIVEN an event B? Cause that's something different.

Anonymous

Anonymous Tue 19 Apr 2022 04:59:10 No.14408519 Report

Quoted By:

>>14408454
even further, if you meant to say
>LOL REAL NUMBERS ARE SETS
this is utterly superfluous to the question. Yes, we all took undergrad analysis. We know about dedekind cuts. You're the person asking if P(A?B) = P(B?A) lmao

Anonymous

Anonymous Tue 19 Apr 2022 05:13:58 No.14408565 Report

Quoted By:

imagine getting filtered by a literal U

Anonymous

Anonymous Tue 19 Apr 2022 05:14:59 No.14408566 Report

Quoted By:

imagine getting filtered by the commutative property of intersection

Anonymous

Anonymous Tue 19 Apr 2022 14:45:13 No.14409333 Report

Quoted By: >>14409357

>>14408465
>Again by definition, P(A?B) = P(B?A) = P(S).
what definition/

Anonymous

Anonymous Tue 19 Apr 2022 14:56:58 No.14409357 Report

Quoted By: >>14409395

>>14409333
The definition of probability function?!?! This is obv troll

Anonymous

Anonymous Tue 19 Apr 2022 15:15:06 No.14409395 Report

Quoted By: >>14409437 >>14409895

>>14409357
I don't see anything in the definition of probability function about all probabilities needing to be equal

Anonymous

Anonymous Tue 19 Apr 2022 15:32:26 No.14409428 Report

Quoted By:

>>14397098
Yes. It's the same. There is no difference. Even if one is conditional on the other, it makes no difference, it's all captured in AND.

It's literally an intersection of sets. Of course A and B is equivalent to B and A.

Now, P(A|B) /= P(B|A), but that's a different problem.

Anonymous

Anonymous Tue 19 Apr 2022 15:36:29 No.14409437 Report

Quoted By:

>>14409395
It's pretty clearly stated desu. The definition of a probability is simply:
P(A) = N(A)/N(U)
Where N(A) is the number of items in the set A and N(U) is the number of items in the set of all outcomes possible for a given event (like rolling dice).

Anonymous

Anonymous Tue 19 Apr 2022 18:26:46 No.14409895 Report

Quoted By:

>>14409395
Ok, let me put this VERY CLEARLY:

Let $U$ be the sample space. Let $A\subseteq U$ and $B\subseteq U$ .

THE DEFINITION OF SET INTERSECTION IS:
$X\cap Y = \{z: z\in X and z\in Y\}.$
SO BY DEFINITION, FOR ANY SETS X AND Y, $X\cap Y = Y\cap X$ . IF YOU WANT TO BE EVEN MORE EXPLICIT, ITS NOT HARD TO SHOW FROM HERE THAT $X\cap Y\subseteq Y\cap X$ AND $Y\cap X\subseteq X\cap Y$ .

Thus, $A\cap B = B\cap A = S\subseteq U$ .

THE FIRST AXIOM FOR PROBABILITIES STATES:

For every subset $E$ of the sample space $U$ , $P(E)\in\mathbb{R}$ and $P(E)\geq 0$ .

BECAUSE $A\cap B = B\cap A = S$ ARE ALL THE SAME SUBSET OF $U$ , THEY MUST BE ASSIGNED TO THE SAME REAL NUMBER BY $P$ .

THAT IS, BY THE DEFINITION OF SET INTERSECTION AND THE FIRST PROBABILITY AXIOM,

$P(A\cap B) = P(B\cap A)$ .

Anonymous

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Anonymous Tue 19 Apr 2022 18:29:42 No.14409904 Report

Quoted By:

>>14397098
This is a who cares.

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