A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let X denote the number of defective pens. Then the variance of X is

Let the random variable $$X$$ be the number of defective pens obtained when 2 pens are drawn without replacement from a box containing 10 pens, out of which 3 are defective.

Because the draws are without replacement, $$X$$ follows a hypergeometric distribution with parameters
    Population size $$N = 10$$
    Number of defectives (successes) $$M = 3$$
    Sample size $$n = 2$$.

The variance formula for a hypergeometric distribution is
$$\text{Var}(X)= n\,\frac{M}{N}\left(1-\frac{M}{N}\right)\frac{N-n}{N-1} \quad -(1)$$

Substitute the given numbers into $$(1)$$:

$$\frac{M}{N} = \frac{3}{10}, \quad 1-\frac{M}{N} = \frac{7}{10}, \quad \frac{N-n}{N-1} = \frac{10-2}{10-1} = \frac{8}{9}.$$

Therefore
$$\text{Var}(X)= 2 \times \frac{3}{10} \times \frac{7}{10} \times \frac{8}{9}$$

Simplify step by step:
$$2 \times \frac{3}{10} = \frac{6}{10},$$
$$\frac{6}{10} \times \frac{7}{10} = \frac{42}{100} = \frac{21}{50},$$
$$\frac{21}{50} \times \frac{8}{9} = \frac{168}{450} = \frac{28}{75}.$$

Hence, the variance of $$X$$ equals $$\dfrac{28}{75}$$.

So the correct choice is Option B.