TITLE: How to prove P(A ∪ B) = P(A) + P(B|A′)P(A′)?
QUESTION [0 upvotes]: What i have tried so far:
given P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
P(A) + P(B) - P(A ∩ B) = P(A ∪ B) = P(A) + P(B|A′)P(A′)
P(B) - P(A ∩ B) = P(B|A′)P(A′)
I dont know how to continue, maybe my approach is entirely wrong :c

REPLY [1 votes]: You have the right idea, it just gets a bit sidetracked in the middle. Your result can be proved as follows:
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$=P(A)+[P(A\cap B)+P(A'\cap B)]-P(A\cap B)$$
$$=P(A)+P(A'\cap B)$$
$$=P(A)+\frac{P(A'\cap B)}{P(A')}P(A')$$
$$=P(A)+P(B\mid A')P(A')$$
{"set_name": "stack_exchange", "score": 0, "question_id": 4324577}