Conditional Probability

danhodge

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Whelp, I have no idea where to put this, and even less of an idea whether anyone here can help me, so here goes nothing:

Studying for my Advanced AI module, and I thought "Hey, let's do some of the questions from the textbook! They can't be much harder...". I then proceeded to cry myself to sleep.

Basically, this question asks:
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And this is the answer they give:
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All I don't understand is, where does the ' A ^ ' go, on the 3rd step? It just disappears and I can't think of any logical reason why :neutral:

Any help will be forever loved and appreciated :angel:
 
There is nothing here, but the question. The links or images you put in are not present.
 
I can see them just fine.

That said, I have no idea. Been a while since I've taken statistics or Logic, Sets, & Proofs :lol:.
 
Good news is I can see the question. Bad news I have forgotten most of that math class. Sorry hopefully someone else can help out.
 
They just performed a simplification, the probability set of A in union with (B in union with A) is the same no matter how many times you say "in union with A", i.e. you can simplify it down to just B union A. That's why they say that bit about "and the fact that A union A == A"
 
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They just performed a simplification, the probability set of A in union with (B in union with A) is the same no matter how many times you say "in union with A"

But isn't the probability of A ^ B the probability that both will happen at the same time? So wouldn't it be the probability of A happening on its own, plus the probability of both happening at the same time? :neutral:

Thanks for helping btw, my lecturer for this module has no idea :lol:
 
But isn't the probability of A ^ B the probability that both will happen at the same time

Yep! Try making two overlapping circles, label A and B. Label the middle overlapping bit A^B

What is the probability that some random point I choose will be in set A^B (the tiny oval in the middle) AND in set A?
Well, set A^B is ENTIRELY within set A. So if it's in set A^B at all, it will 100% be in set A as well. We don't need to calculate that.

Or to state it more clearly, if we have P ( A ^ ( A ^ B)) then the associative law says we can also rewrite that as P((A ^ A) ^ B)
P(A ^ A) = P(A)


Also nice! Which AI course are you doing?
 
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