Q42 / 75 · JEE Main 2025 (4 Apr Shift 1) · Mathematics
A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let $X$ denote the number of defective pens. Then the variance of X is
- A$\frac{11}{15}$
- Correct answerB$\frac{28}{75}$
- C$\frac{2}{15}$
- D$\frac{3}{5}$
Solution
\begin{array}{|r|c|c|c|}\hline x & \mathrm{x}=0 & \mathrm{x}=1 & \mathrm{x}=2 \\\hline \mathrm{P}(\mathrm{x}) & \frac{{ }^7 \mathrm{C}_2}{{ }^{10} \mathrm{C}_2} & \frac{{ }^7 \mathrm{C}_1^3 \mathrm{C}_1}{{ }^{10} \mathrm{C}_2} & { }^3{ }^{10} \mathrm{C}_2 \\\hline\end{array}
$\begin{aligned} & \mu=\Sigma \mathrm{x}_{\mathrm{i}} \mathrm{P}\left(\mathrm{x}_{\mathrm{i}}\right)=0+\frac{7}{15}+\frac{2}{15}=\frac{3}{5} \\ & \text { Variance }(\mathrm{x})=\Sigma \mathrm{P}_{\mathrm{i}}\left(\mathrm{x}_{\mathrm{i}}-\mu\right)^2=\frac{28}{75}\end{aligned}$
$\begin{aligned} & \mu=\Sigma \mathrm{x}_{\mathrm{i}} \mathrm{P}\left(\mathrm{x}_{\mathrm{i}}\right)=0+\frac{7}{15}+\frac{2}{15}=\frac{3}{5} \\ & \text { Variance }(\mathrm{x})=\Sigma \mathrm{P}_{\mathrm{i}}\left(\mathrm{x}_{\mathrm{i}}-\mu\right)^2=\frac{28}{75}\end{aligned}$
Hence, the correct answer is option: 'B'.