Given a lot of 12 items, 3 are defective.

Good items, $$12-3=9$$

Let $$X$$ denote the number of defective items.

So, value of $$X=0,1,2,3$$

A sample of $$S$$ items is drawn.

$$P(X=0)=G G G G G$$

(here $$G$$ is good item and $$d$$ is defective)

$$\begin{aligned} & \frac{9}{12} \cdot \frac{8}{11} \cdot \frac{7}{10} \cdot \frac{6}{9} \cdot \frac{5}{8}=\frac{21}{132}=\frac{7}{44} \\ & P(X=1)=5\left[\frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 3}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}\right]=\frac{21}{44} \\ & P(X=2)=5\left[\frac{9 \cdot 8 \cdot 7 \cdot 3 \cdot 2}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}\right]=\frac{14}{44} \\ & P(X=3)=5\left[\frac{3 \cdot 2 \cdot 1 \cdot 9 \cdot 8}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}\right]=\frac{2}{44} \\ & P(X=4)=0 \\ & P(X=5)=0 \end{aligned}$$

$$\begin{aligned} & \sigma_x^2=\sum X^2 P(x)-\left(\sum x P(x)\right)^2 \\ & =\frac{95}{44}-\left(\frac{55}{44}\right)^2 \\ & =\frac{4180-3025}{1936}=\frac{1155}{1936}=\frac{105}{176}=\frac{m}{n} \\ & =n-m=71 \end{aligned}$$