1. ## Conditional Probability Proof

Hello,

I am new here! Nice forum by the way.

So i have a weird proof to solve there that kind of remind this one but its seems that its quite different, after all.

B, C are mutually exclusive.

And i have to proove this:

P(A | B ∪ C) = [ P(B) * P(A | B) / P(B) + P(C) ] + [ P(C) * P(A | C) / P(B) + P(C) ]

I have tried load of things, but mostly i am doing cycles around. I am definatelly missing some keypoint ?

Shoud i try proove it by taking the long part and reach P(A | B ∪ C) , or should i take small part reach the big one?

Honestly i have tryed both way and nothing absolutely nothing.

On my best attempt i reached this point (by taking the long part trying to reach the 1st part):

P(A∩B) + P(A∩C) / P(B) + P(C) and i cant continue from this point. i mean that if "+" was like "or" i could probably say that

P(A∩B) + P(A∩C) / P(B) + P(C) = P(A∩(B∪C)) / P(B) +P(C) and finish it here (but its wrong)...

Any ideas there ?

2. ## Re: Conditional Probability Proof

Originally Posted by Yonas
And i have to proove this:

P(A | B ∪ C) = [ P(B) * P(A | B) / P(B) + P(C) ] + [ P(C) * P(A | C) / P(B) + P(C) ]
You do know that in the standard order or operations division comes before addition, right?

Originally Posted by Yonas
On my best attempt i reached this point (by taking the long part trying to reach the 1st part):

(P(A∩B) + P(A∩C)) / (P(B) + P(C)) and i cant continue from this point. i mean that if "+" was like "or" i could probably say that

(P(A∩B) + P(A∩C)) / (P(B) + P(C)) = P(A∩(B∪C)) / (P(B) +P(C)) and finish it here (but its wrong)...
It's not wrong. First, since B and C are mutually exclusive (i.e., are disjoint subsets of the sample space), we have P(B) + P(C) = P(B ∪ C). Second, for the same reason A ∩ B and A ∩ C are also disjoint, so P(A ∩ B) + P(A ∩ C) = P((A ∩ B) ∪ (A ∩ C)) = P(A ∩ (B ∪ C)).

3. ## Re: Conditional Probability Proof

Originally Posted by emakarov
You do know that in the standard order or operations division comes before addition, right?
Yes mate, i know that, and its quite trashy from me not using parentheses, excuse me.

Originally Posted by emakarov
It's not wrong. First, since B and C are mutually exclusive (i.e., are disjoint subsets of the sample space), we have P(B) + P(C) = P(B ∪ C). Second, for the same reason A ∩ B and A ∩ C are also disjoint, so P(A ∩ B) + P(A ∩ C) = P((A ∩ B) ∪ (A ∩ C)) = P(A ∩ (B ∪ C)).
Oh, i see! Quite enlighting answer pal. Thank you a lot.

So, do you think my way to proove going from [P(B) * P(A | B) / ( P(B) + P(C) ) ] + [ P(C) * P(A | C) / ( P(B) + P(C) ) ] to P(A | B ∪ C) is right, just like i described it later? I mean, do i really proove it like this, it seems that i really do, but i cant be sure like 100% , i am not very familiar with this field. Can you call this "legitimate" proof ?

Thanks

4. ## Re: Conditional Probability Proof

Ultimately, it does not matter whether you prove equality by converting the left-hand side into the right-hand side or vice versa. It may be easier to come up with a transformation in one direction, but once the chain of equalities is written, it can be read both ways.

If you'd like, post a complete proof for a feedback.