QJEEM24Ma0c1c: JEE Main 2024 Mathematics - Class 12 JEEM24Ma0c1c Solution: From a lot of 12 items containing 3 defectives, a sample of 5…

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)

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

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