Three defective oranges are accidently mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If $$x$$ denote the number of defective oranges, then the variance of $$x$$ is

We need to find the variance of $$x$$, where $$x$$ is the number of defective oranges when 2 oranges are drawn from a lot of 3 defective and 7 good oranges.

For a hypergeometric distribution (sampling without replacement), the variance is given by $$\text{Var}(x) = E(x^2) - [E(x)]^2$$. Alternatively, we can compute it directly from the probability distribution of $$x$$.

First, we find the probability distribution of $$x$$. The total number of ways to choose 2 from 10 is $$\binom{10}{2} = 45$$.

For $$x = 0$$, both oranges are good, so $$P(x=0) = \frac{\binom{7}{2}}{\binom{10}{2}} = \frac{21}{45} = \frac{7}{15}$$.

For $$x = 1$$, one defective and one good, so $$P(x=1) = \frac{\binom{3}{1}\binom{7}{1}}{\binom{10}{2}} = \frac{3 \times 7}{45} = \frac{21}{45} = \frac{7}{15}$$.

For $$x = 2$$, both defective, so $$P(x=2) = \frac{\binom{3}{2}}{\binom{10}{2}} = \frac{3}{45} = \frac{1}{15}$$.

Next, the expected value of $$x$$ is calculated as $$E(x) = 0 \times \frac{7}{15} + 1 \times \frac{7}{15} + 2 \times \frac{1}{15} = 0 + \frac{7}{15} + \frac{2}{15} = \frac{9}{15} = \frac{3}{5}$$.

Then, we compute $$E(x^2) = 0^2 \times \frac{7}{15} + 1^2 \times \frac{7}{15} + 2^2 \times \frac{1}{15} = 0 + \frac{7}{15} + \frac{4}{15} = \frac{11}{15}$$.

Finally, substituting these values into the variance formula gives $$\text{Var}(x) = E(x^2) - [E(x)]^2 = \frac{11}{15} - \left(\frac{3}{5}\right)^2 = \frac{11}{15} - \frac{9}{25} = \frac{11 \times 5}{75} - \frac{9 \times 3}{75} = \frac{55}{75} - \frac{27}{75} = \frac{28}{75}$$.

The correct answer is Option (1): 28/75.