Like if I wanted to search for pictures with "Blue_eyes" or "blood" but if I was to search for "Blue_eyes blood" I would only get pictures contaning blue eyes AND blood. Is there a way to search for one thing or another?

Maskarade said:
Like if I wanted to search for pictures with "Blue_eyes" or "blood" but if I was to search for "Blue_eyes blood" I would only get pictures contaning blue eyes AND blood. Is there a way to search for one thing or another?

... what?

EDIT: Okay, here's what to do. Delete System32 and do a barrel roll.

You can use ~ in front of at least two tags to search for posts that contain at least one of those tags. So in your case, try ~blue_eyes ~blood. That should return all posts with blue_eyes and/or blood.

You can combine that with other tags, so ~blue_eyes ~blood fox would return posts that have fox AND either blue_eyes or blood (or both).

There's no way to have two different 'or' searches though, so you can't do something like (fox or raccoon) AND (blue_hair or red_hair).

Maskarade said:
Like if I wanted to search for pictures with "Blue_eyes" or "blood" but if I was to search for "Blue_eyes blood" I would only get pictures contaning blue eyes AND blood. Is there a way to search for one thing or another?

~blue_eyes ~blood

~blue_eyes ~white_dragon

If P(X)=0.245 and P(X ∩ Y)=0.655, what does P(Y|X) equal?

male -female
shows only males
https://e621.net/post?tags=male+-female

Don't forget to add a space and if you don't want to see a particular thing, a -(minus) symbol in front.

male big_cock -female <3

What about Z?
Where's the love for my vertices?

Aeshma said:
What about Z?
Where's the love for my vertices?

They just don't care about vertices anymore. =>.<=

Furry_Fanatic said:
If P(X)=0.245 and P(X ∩ Y)=0.655

Umm... it can't? P(X ∩ Y) is the set containing both X and Y, or the joint probability of both X and Y occurring together, ergo it will always be a subset of P(X) and the probability of occurrence will be less (unless you allow probabilities greater than 1, which is nonsensical). Taking your values at their face, P(Y|X) would have come out to ~2.67, which is patently wrong in probability.

Perhaps you meant P(X ∩ Y)=0.245 and P(X)=0.655, at which point...

Furry_Fanatic said:
what does P(Y|X) equal?

...we plug our terms into the formula P(Y|X)=P(X ∩ Y)/P(X) and discover P(Y|X)=~0.374.

Now, if instead you meant P(X)=0.245 and P(X ∪ Y)=0.655, then we have two options, either X and Y are mutually exclusive, or they aren't.

If they are mutually exclusive, we simply know that P(Y|X)=0. We can observe from P(X ∪ Y)=P(X)+P(Y) that P(Y)=0.41, but that's not the question being asked.

If they are not mutually exclusive, we can only solve the problem if they are independent, because we don't otherwise know P(X ∩ Y). Since P(Y|X)=P(X ∩ Y)/P(X), we can observe that this becomes P(Y|X)=P(X)P(Y)/P(X), or P(Y|X)=P(Y). (Intuitively, if they are independent, then the odds of Y occurring are the same regardless of whether X has occurred.) Now, what is P(Y)? We know P(X ∪ Y)=0.655, ergo the complement P(not X ∩ not Y)=P(not X)P(not Y)=0.345. Therefor P(not Y)=P(not X ∩ not Y)/P(not X). Since the complement P(not X)=1-P(X) or 0.755, and we know the complement P(not X ∩ not Y)=0.345, we can determine that P(not Y)=0.345/0.755=~0.457. P(Y) is therefor ~0.543, as is P(Y|X).

ikdind said:
blah blah blah...

It's penis... >:I

Xch3l said:
It's penis... >:I

You mean this.

ikdind said:
Statistics

I just threw in some values, man. You took it to a whole new level. It was only intended to be a joke, but if you wanted to see it through fine by me.

Furry_Fanatic said:
I just threw in some values, man. You took it to a whole new level. It was only intended to be a joke, but if you wanted to see it through fine by me.

:P

I take everything too seriously.

ikdind said:
:P

I take everything too seriously.

ikdind is pregnant.

ikdind said:
:P

I take everything too seriously.

You want to do some inferential tests next? I'll make sure that you check those assumptions and conditions.

furballs_dc said:
ikdind is pregnant.

Merriam-Webster definition 1: archaic: cogent.
I am having power to compel or constrain? Nobody told me this! I need to head to a bar, there are people there I want free drinks and... other things... from.

Just as planned...