If A and B are two events such that

If $A \subset B$, then $A \cap B=A$

$\Rightarrow P(A \cap B)=P(A)$

Also, $P(A)

Consider $P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{P(A)}{P(B)} \neq \frac{P(B)}{P(A)} \ldots$(1)    

Consider $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{\mathrm{P}(\mathrm{A})}{\mathrm{P}(\mathrm{B})}$  ...(2)

It is known that, P (B) ≤ 1

$\Rightarrow \frac{1}{P(B)} \geq 1$

$\Rightarrow \frac{P(A)}{P(B)} \geq P(A)$

From $(2)$, we obtain

$\Rightarrow \mathrm{P}(\mathrm{A} \mid \mathrm{B}) \geq \mathrm{P}(\mathrm{A})$              ...(3)

$\therefore \mathrm{P}(\mathrm{A} \mid \mathrm{B})$ is not less than $\mathrm{P}(\mathrm{A})$

Thus, from (3), it can be concluded that the relation given in alternative C is correct.