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 ?
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 ?
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.