Why can we expand P(A) as P(A/B)P(B) + P(A/B')P(B')?

[Avichal]

P(B/A) = P(A/B).P(B) / P(A)

Later we expand P(A) as P(A/B).P(B) + P(A/B).P(B) ... B is complement of B

I don't understand how we can expand P(A) like that. Doesn't that assume that A ℂ B?

 

[marcusl]

Think about what it is saying. The probability that A happens is the probability that A happens given that B happens plus the probability that A happens given that B doesn't happen. Both cases are needed to cover all possibilities.

 

[Avichal]

Well basically what my book says is that : -
P(A) = P(A/B₁).P(B₁) + P(A/B₂).P(B₂) + ... + P(A/B_n).P(B_n)

Doesn't this assume that B₁ U B₂ ... U B_n is a super-set of A?

 

[D H]

Doesn't this assume that B₁ U B₂ ... U B_n is a super-set of A?

Of course.

Your text should have specified that B₁, B₂, ···, B_n are a set of mutually disjoint subsets of the universe U of possible outcomes and that B₁ ∪ B₂ ∪ ··· ∪ B_n=U. The set A must be a subset of this universe of outcomes U; otherwise it doesn't even make sense to talk about P(B₁|A).

 

[Avichal]

Of course.

Your text should have specified that B₁, B₂, ···, B_n are a set of mutually disjoint subsets of the universe U of possible outcomes and that B₁ ∪ B₂ ∪ ··· ∪ B_n=U. The set A must be a subset of this universe of outcomes U; otherwise it doesn't even make sense to talk about P(B₁|A).

It didn't. Anyways, thank you. This clears my doubt.