Prove that $P(A|B) + P(A|B^{c}) = P(A)$

Question

This is for homework, so i just need a hint.

Thanks in advice.

Answer 1

It's not true, so you can't prove it.

For example, consider the case when $A$ and $B$ are the same event, and $P(B) = 0.5$. In that case, $P(A|B) = 1$, $P(A|B^c) = 0$, and $P(A) = 0.5 \neq P(A|B) + P(A|B^c)$.

Answer 2

This equality is not valid. I think you're forgetting the factors $P(B)$ and $P(B^c)$ for $P(A|B^c)$ and $P(A| B)$.

$$ \begin{array}{rll} P(A)=& P(A\cap \Omega) & \mbox{ by }A\cap \Omega=A\\ =& P(A\cap(B^c\cup B))& \mbox{ by }\Omega=B^c\cup B\\ =& P((A\cap B^c)\cup (A\cap B))& \mbox{ by distributive propertie }\\ =& P(A\cap B^c)+ P(A\cap B)& \mbox{ by cause } (A\cap B^c)\cap (A\cap B)=\emptyset\\ =& \dfrac{P(A\cap B^c)}{P(B^c)}\cdot P(B^c) + \dfrac{P(A\cap B)}{P(B)}\cdot P(B)& \mbox{ If } P(B)\neq 0 \mbox{ and } P(B^c)\neq 0\\ =& P(A|B^c)\cdot P(B^c) + P(A| B)\cdot P(B)& \mbox{ If } P(B)\neq 0 \mbox{ and } P(B^c)\neq 0\\ \end{array} $$