JEE Main 2024 — Probability Question with Solution

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.

$$P(X=0)=GGGGG$$

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

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

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