Answer : (A) 𝑃(𝐵 | 𝐴) ≥ 𝑃(𝐵)
Given $[B \subset A ]\rightarrow [P(A\cap B) = P(B) ]$ , Conditional Probability :
so,
$P(\frac{A}{B})$ = $\frac{P(A \cap B)}{P(B)}$ = $\frac{P(B)}{P(B)} = 1$
and
$P(\frac{B}{A})$ = $\frac{P(A \cap B)}{P(A)}$ = $\frac{P(B)}{P(A)}$
since $ 0 < P(B) \leq P(A) \leq 1 $,
when some number P(B) ( $\leq 1 )$ divided by another number greater than that number P(A) ( $ P(B) \leq P(A) \leq 1 $ )
$\implies \frac{P(B)}{P(A)} \geq P(B) \implies P(\frac{B} {A}) \geq P(B) $