"Suppose $A\subseteq B$ and $Pr(B)\gt0$, show that $Pr(A|B)\geq Pr(A)$." (There are no other antecedents given.)
$Pr(A|B)=\frac{Pr(A)}{Pr(B)}$. Which implies, $Pr(A|B)Pr(B)=Pr(A)$. I don't think the inequality is true. I think the inequality should have been the other way around. Am I missing something?
$Pr(A|B)=\frac {\Pr(A \cap B)} {Pr(B)}=\frac {Pr(A)} {Pr(B)} \geq Pr(A)$.
The first equality is the definition. The second equality follows from the fact that $A \subseteq B$ which implies $A \cap B=A$. The ineqility at the end follows from the fact that $P(B)\leq 1$.