1. Define the random variable

Let $X$ be the number of defective oranges when $2$ oranges are drawn from the $10$ oranges.

There are:

• $3$ defective oranges
• $7$ good oranges
• total $10$ oranges

Since two oranges are drawn without replacement, $X$ can take values: $X=0,1,2$

This is a hypergeometric distribution.

2. Find the probability distribution of $X$

Total number of ways to draw $2$ oranges from $10$: $\binom{10}{2}=45$

Case 1: $X=0$

Both drawn are good: $P(X=0)=\frac{\binom{7}{2}}{\binom{10}{2}}=\frac{21}{45}=\frac{7}{15}$

Case 2: $X=1$

One defective and one good: $P(X=1)=\frac{\binom{3}{1}\binom{7}{1}}{\binom{10}{2}}=\frac{21}{45}=\frac{7}{15}$

Case 3: $X=2$

Both drawn are defective: $P(X=2)=\frac{\binom{3}{2}}{\binom{10}{2}}=\frac{3}{45}=\frac{1}{15}$

Check: $\frac{7}{15}+\frac{7}{15}+\frac{1}{15}=1$

3. Compute the mean $E(X)$

$E(X)=\sum xP(X=x)$

$E(X)=0\cdot \frac{7}{15}+1\cdot \frac{7}{15}+2\cdot \frac{1}{15}$

$E(X)=\frac{7}{15}+\frac{2}{15}=\frac{9}{15}=\frac{3}{5}$

4. Compute $E(X^2)$

$E(X^2)=\sum x^2P(X=x)$

$E(X^2)=0^2\cdot \frac{7}{15}+1^2\cdot \frac{7}{15}+2^2\cdot \frac{1}{15}$

$E(X^2)=\frac{7}{15}+\frac{4}{15}=\frac{11}{15}$

5. Compute the variance

$\operatorname{Var}(X)=E(X^2)-[E(X)]^2$

$\operatorname{Var}(X)=\frac{11}{15}-\left(\frac{3}{5}\right)^2$

$\operatorname{Var}(X)=\frac{11}{15}-\frac{9}{25}$

Taking LCM $75$: $\frac{11}{15}=\frac{55}{75},\qquad \frac{9}{25}=\frac{27}{75}$

So, $\operatorname{Var}(X)=\frac{55}{75}-\frac{27}{75}=\frac{28}{75}$

6. Match with the options

$\boxed{\operatorname{Var}(X)=\frac{28}{75}}$

Hence the correct option is:

D: $\dfrac{28}{75}$