# Math Help - conditional probability w/ boolean operators

1. ## 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!

2. I am not sure that I understand the question.
But this is true: $\left( {A \cup B \cup C} \right) \cap \left( {A \cup B} \right) = \left( {A \cup B} \right)$.

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

3. 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....

4. It you want to use symbols it makes is easy to read.
$$P(A\cap B\cap C)| A\cap B)$$ you will get $P(A\cap B\cap C)| A\cap B)$.

Here are some more.
$$\cap$$ gives $\cap$ is intersection
$$\subseteq$$ gives $\subseteq$ is subset
$$\cup$$ gives $\cup$ is union
$$\in$$ gives $\in$ is element
$$\emptyset$$ gives $\emptyset$ is emptyset

5. sorry i seem so slow. this is self-learning....
$(A\cap B\cap C) \cap (A \cap B) = (A\cap B)$ 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?

6. Originally Posted by uneakbreed
sorry i seem so slow. this is self-learning....
$(A\cap B\cap C) \cap (A \cap B) = (A\cap B)$ 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 $(A\cap B\cap C) \cap (A \cap B) = (A\cap B\cap C)$.

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

7. 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
$P(A|B) = \frac{P(A \cap B) \cap P(B)}{P(B)}$

then why not multiply $P(A \cap B \cap C)$ to itself? In other words, if $P(A \cap B) =P(A \cap B \cap C)$ making the equation be

$P(A \cap B \cap C) | A \cap B) = \frac{P(A \cap B \cap C) \cap P(A \cap B \cap C)}{P(A \cap B)}$

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....

8. Originally Posted by uneakbreed
$P(A \cap B \cap C) | A \cap B) = \frac{P(A \cap B \cap C) \cap P(A \cap B \cap C)}{P(A \cap B)}$
Given that all we have is this
$P(A \cap B \cap C) | A \cap B) = \frac{P(A \cap B \cap C) }{P(A \cap B)}$
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