A box contains $6$ pens, $2$ of which are defective. Two pens are taken randomly from the box. If random variable $x:$ Number of defective pens obtained, then standard deviation of $x=$

\(x\) : no. of defective pens
Two pens are taken from box
\(\therefore x\) can taken values \(0,1,2\)
\(\begin{array}{l}p(x=0)=\frac{{ }^4 C_2}{{ }^6 C_2}=\frac{4 \times 3}{6 \times 5}=\frac{2}{5}=\frac{6}{15} \\ p(x=1)=\frac{{ }^2 C_1 \times{ }^4 C_1}{{ }^6 C_2}=\frac{2 \times 4 \times 2 \times 1}{6 \times 5}=\frac{8}{15} \\ p(x=2)=\frac{{ }^2 C_2}{{ }^6 C_2}=\frac{1 \times 2 \times 1}{6 \times 5}=\frac{1}{15}\end{array}\)
\(\begin{array}{||l|l|l|l||} \hline \hline X & P & x_i p_i & x_i^2 p_i \\ \hline \hline 0 & \frac{6}{15} & 0 & 0 \\ \hline 1 & \frac{8}{15} & \frac{8}{15} & \frac{8}{15} \\ \hline 2 & \frac{1}{15} & \frac{2}{15} & \frac{4}{15} \\ \hline \hline \end{array}\)
\(E(x)=\frac{10}{15}\)
\(=\frac{2}{3}\)
\(E\left(x^2\right)=\frac{12}{15}\)
\(=\frac{4}{5}\)
Standard deviation \(=\sqrt{E\left(x^2\right)-[E(x)]^2}\)
Standard deviation \(=\sqrt{\left(\frac{4}{5}\right)-\left(\frac{2}{3}\right)^2}\)
\(\begin{array}{l}=\sqrt{\frac{4}{5}-\frac{4}{9}} \\ =\sqrt{\frac{4 \times 4}{45}} \\ =\frac{4}{3 \sqrt{5}}\end{array}\)