"From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of $X$ is $\frac{m}{n}$, where $\operatorname{gcd}(m, n)=1$, then $n-m$ is equal to _________."

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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)

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

$X$	0	1	2	3
$P(X)$	$ \frac{7}{44} $	$ \frac{21}{44} $	$ \frac{14}{44} $	$ \frac{2}{44} $
$XP(X)$	0	$ \frac{21}{44} $	$ \frac{28}{44} $	$ \frac{6}{44} $
$X^2P(X)$	0	$ \frac{21}{44} $	$ \frac{56}{44} $	$ \frac{18}{44} $

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

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