1. Interpret the notation:

   $P\left(\frac{C}{D}\right) = P(C\mid D)$

   i.e. the conditional probability of $C$ given $D$.

2. Use the definition of conditional probability:

   $P(C\mid D)=\frac{P(C\cap D)}{P(D)}$

   given that $P(D)\ne 0$.

3. Since $C\subset D$, whenever $C$ occurs, $D$ also occurs. Therefore,

   $C\cap D = C$

   so

   $P(C\mid D)=\frac{P(C)}{P(D)}.$

4. Now compare $P(C\mid D)$ with $P(C)$.

   Since $C\subset D$,

   $P(C)\le P(D).$

   Also, $0 < P(D) \le 1$, hence

   $\frac{1}{P(D)} \ge 1.$

   Multiplying by $P(C)\ge 0$,

   $\frac{P(C)}{P(D)} \ge P(C).$

   Therefore,

   $P(C\mid D) \ge P(C).$

5. Check the options:

   • A: $P\left(\frac{C}{D}\right) \ge P(C)$ ✅ correct
   • B: $P\left(\frac{C}{D}\right) < P(C)$ ❌ false
   • C: $P\left(\frac{C}{D}\right) = \frac{P(D)}{P(C)}$ ❌ false, actual formula is $\frac{P(C)}{P(D)}$
   • D: $P\left(\frac{C}{D}\right) = P(C)$ ❌ not always true; equality only when $P(D)=1$ or $P(C)=0$

Hence the correct option is A.