Thread: conditional probability w/ boolean operators

Hi, Can someone help me out? I just want to make sure i'm right since the solution doesn't spell out each step....

As far as my understanding p(a | b) = [p(a /\ b)]/ p(b)

but, i'm looking at problems that have
p(a /\ b /\ c | p(a /\ b) = [p(a /\ b /\ c)] / p(a /\ b)

or
p(a V b V c | p(a V b) = [p(a V b V c)] / p(a V b)


is this because of the identity law? no need to include p(a /\ b) in the numerator because they are already included in the first clause? in other words,
p(a /\ b /\ c)] /\ p(a /\ b) is redundant and can be reduced to p(a /\ b /\ c) -as is the same for p(a V b V c) /\ p(a V b V c) ?

Can someone confirm this? just wanna make sure i'm not misunderstanding the solution step... thanks!

I am not sure that I understand the question.
But this is true: .

Now I don't think that is what you have. Is it?

yeah i'm still trying to get a handle on the cool math app you guys use so my fault for not using it yet.
mainly the solution has p(a /\ b /\ c | p(a /\ b) which could be rewritten a p(x | y). and p(x|y) = p(x /\ y) / p(y)

That being said, if If x =a /\ b /\ c and y = a /\ b then shouldn't the equation be:
[p(a /\ b /\ c)] /\ p(a /\ b) ] / p(a /\ b) ?

But, the solution has it as:
[p(a /\ b /\ c)] / p(a /\ b) instead....

It you want to use symbols it makes is easy to read.
[tex]P(A\cap B\cap C)| A\cap B) [/tex] you will get .

Here are some more.
[tex] \cap [/tex] gives is intersection
[tex] \subseteq [/tex] gives is subset
[tex] \cup [/tex] gives is union
[tex] \in [/tex] gives is element
[tex] \emptyset [/tex] gives is emptyset

sorry i seem so slow. this is self-learning....
seems like a mystery to me.

i'm looking at the distributive law and the absorption law as those that most closely relate but i can't see either applying in this case. sorry, but can you spell it out for this layman?

Originally Posted by uneakbreed

sorry i seem so slow. this is self-learning....
seems like a mystery to me. i'm looking at the distributive law and the absorption law as those that most closely relate but i can't see either applying in this case. sorry, but can you spell it out for this layman?

I see why it is mystery to you: IT IS WRONG.
It should be .

Sorry about that. But I was working off your OP that used union rather than intersection.

AHHH! ok. so it is the identity law. cool.

so since they are equal, why don't i need to multiply it against itself? in other words, although they are equal, since the original equation is


then why not multiply to itself? In other words, if making the equation be



also:
is there a web resource that you use that lists all the identities of boolean algebra? i found a couple but i never know how complete they are....

Originally Posted by uneakbreed

Given that all we have is this

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