How $\mathbb{P} (A | B \cup B^c)$ is $\mathbb{P} (A | B) \cdot P(B) + \mathbb{P} (A | B^c) \cdot \mathbb{P} (B^c)$?

Question

P(A) = $P(A|\Omega)$

        = $ P(A|B \cup B^c)$

But how to reach P(A | B) * P(B) + P(A | B$^c$) * P(B$^c$) from P(A | B U B$^c$) ?

Answer 1

Quickly, $\mathsf P(A\mid B\cup B^\complement)=\mathsf P(A)$ because $B\cup B^\complement$ is the entire outcome set.

Slowly, $$\begin{align}\mathsf P(A\mid B\cup B^\complement)&=\dfrac{\mathsf P(A\cap( B\cup B^\complement))}{\mathsf P(B\cup B^\complement)}\\&=\dfrac{\mathsf P(A\cap B)+\mathsf P(A\cap B^\complement)}{1}\\&=\mathsf P(A\mid B)~\mathsf P(B)+\mathsf P(A\mid B^\complement)~\mathsf P(B^\complement)\end{align}$$