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(B) The total number of ways to choose $2$ pens from $10$ is $^{10}C_2 = \frac{10 	imes 9}{2} = 45$.The random variable $X$ can take values $0, 1, 2$

Vedclass · Accepted Answer

$\frac{28}{75}$

Vedclass · Answer

(B) The total number of ways to choose $2$ pens from $10$ is $^{10}C_2 = \frac{10 	imes 9}{2} = 45$.The random variable $X$ can take values $0, 1, 2$.The probability distribution is:$X$$0$$1$$2$$P(X)$$\frac{^7C_2}{45} = \frac{21}{45}$$\frac{^7C_1 	imes ^3C_1}{45} = \frac{21}{45}$$\frac{^3C_2}{45} = \frac{3}{45}$Mean $E(X) = \sum x_i P(x_i) = 0 	imes \frac{21}{45} + 1 	imes \frac{21}{45} + 2 	imes \frac{3}{45} = \frac{21+6}{45} = \frac{27}{45} = \frac{3}{5}$.Expectation $E(X^2) = \sum x_i^2 P(x_i) = 0^2 	imes \frac{21}{45} + 1^2 	imes \frac{21}{45} + 2^2 	imes \frac{3}{45} = \frac{21+12}{45} = \frac{33}{45} = \frac{11}{15}$.Variance $Var(X) = E(X^2) - [E(X)]^2 = \frac{11}{15} - (\frac{3}{5})^2 = \frac{11}{15} - \frac{9}{25} = \frac{55-27}{75} = \frac{28}{75}$.

Outcome	$\omega_{1}$	$\omega_{2}$	$\omega_{3}$	$\omega_{4}$	$\omega_{5}$	$\omega_{6}$	$\omega_{7}$
Probability	$-0.1$	$0.2$	$0.3$	$0.4$	$-0.2$	$0.1$	$0.3$

$X$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X)$	$0.1$	$0.2$	$0.2$	$0.3$	$0.15$	$0.05$

$A$ 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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