From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $$X$$ denote the number of defective items in the sample. If the variance of $$X$$ is $$\sigma^2$$, then $$96\sigma^2$$ is equal to ______

Hypergeometric distribution: $$X$$ = number of defective items in sample of 5 from lot of 10 (3 defective).

Mean: $$E(X) = n \cdot ?\frac{K}{N} = 5 \cdot ?\frac{3}{10} = ?\frac{3}{2}$$

Variance: $$\sigma^2 = n \cdot ?\frac{K}{N} \cdot ?\frac{N-K}{N} \cdot ?\frac{N-n}{N-1} = 5 \cdot ?\frac{3}{10} \cdot ?\frac{7}{10} \cdot ?\frac{5}{9} = ?\frac{525}{900} = ?\frac{7}{12}$$

$$96\sigma^2 = 96 \times ?\frac{7}{12} = 56$$.

The answer is 56.