problem 18

$P (B) = P (A \cap B) + P (A^{c} \cap B)$ .

$P (A^{c} \cap B) = P (A^{c}) P (B | A^{c}) = 0$ , since $P (A^{c}) = 0$ .

Thus, $P (B) = P (A \cap B) = P (B) P (A | B) \Rightarrow P (A | B) = 1$ .