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

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Aicharya · Accepted Answer

$\begin{array}{|l|l|l|l|} \hline x & \mathrm{x}=0 & \mathrm{x}=1 & \mathrm{x}=2 \ \hline P(x) & \frac{{ }^7 C_2}{{ }^{10} C_2} & \frac{{ }^7 C_1{ }^3 C_1}{{ }^{10} C_2} & \frac{{ }^3 C_2}{{ }^{10} C_2} \ \hline \end{array}$ $\begin{aligned} & \mu=\Sigma \mathrm{x}_{\mathrm{i}} \mathrm{P}\left(\mathrm{x}_{\mathrm{i}}\right)=0+\frac{7}{15}+\frac{2}{15}=\frac{3}{5} \ & \text { Variance }(\mathrm{x})=\Sigma \mathrm{P}_{\mathrm{i}}\left(\mathrm{x}_{\mathrm{i}}-\mu\right)^2=\frac{28}{75}\end{aligned}$ ,

A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let $X$ denote…

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