Prove : $ P(B|A) + P(B|A') =1$

Question

I came across this question in my test and could not solve it. Do I have to prove it using $P(B|A) + P(B'|A) = 1?$

Note: $B' = B^c$

Answer 1

I misinterpreted your question, here is a proof for the true statement $P(B|A) + P(B'|A) =1$ :
$P(B|A) = \dfrac{P(B\cap A)}{P(A)}$, $P(B'|A) = \dfrac{P(B'\cap A)}{P(A)}$

Now sum it up to get

$P(B|A) +P(B'|A) = \dfrac{P(B\cap A)}{P(A)}+ \dfrac{P(B'\cap A)}{P(A)} = \dfrac{P(B\cap A)+P(B'\cap A)}{P(A)}$

Now, what is $P(B\cap A)+P(B'\cap A) $ ?

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Now, $P(A|B) + P(A|B') =1$ is false, for example : $\Omega=\{1,2,3\}$ with uniform probability.

Take $A=\{1\} B=\{1,2\}$ then $P(A|B)=1/2$ and $P(A|B')=0$ and the sum is not $1$.