A box contains 6 pens,2 of which are defectiv | vedclass

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(D) $x$ : number of defective pens. Two pens are taken from the box. Therefore,$x$ can take values $0, 1, 2$. $P(x=0) = \frac{{}^4C_2}{{}^6C_2} = \fra

Vedclass · Accepted Answer

$\frac{4}{3 \sqrt{5}}$

Vedclass · Answer

(D) $x$ : number of defective pens. Two pens are taken from the box. Therefore,$x$ can take values $0, 1, 2$. $P(x=0) = \frac{{}^4C_2}{{}^6C_2} = \frac{6}{15}$ $P(x=1) = \frac{{}^2C_1 	imes {}^4C_1}{{}^6C_2} = \frac{8}{15}$ $P(x=2) = \frac{{}^2C_2}{{}^6C_2} = \frac{1}{15}$   $x_i$ $P(x_i)$ $x_i P(x_i)$ $x_i^2 P(x_i)$   $0$ $\frac{6}{15}$ $0$ $0$   $1$ $\frac{8}{15}$ $\frac{8}{15}$ $\frac{8}{15}$   $2$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{4}{15}$   $E(x) = \sum x_i P(x_i) = 0 + \frac{8}{15} + \frac{2}{15} = \frac{10}{15} = \frac{2}{3}$ $E(x^2) = \sum x_i^2 P(x_i) = 0 + \frac{8}{15} + \frac{4}{15} = \frac{12}{15} = \frac{4}{5}$ $	ext{Variance}(x) = E(x^2) - [E(x)]^2 = \frac{4}{5} - (\frac{2}{3})^2 = \frac{4}{5} - \frac{4}{9} = \frac{36-20}{45} = \frac{16}{45}$ $	ext{Standard Deviation} = \sqrt{	ext{Variance}(x)} = \sqrt{\frac{16}{45}} = \frac{4}{3 \sqrt{5}}$

$x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X=x)$	$a$	$a$	$a$	$b$	$b$	$0.3$

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

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