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(D) Total apples = $3 + 15 = 18$.We draw $2$ apples. The total number of ways to choose $2$ apples is $^{18}C_2 = \frac{18 	imes 17}{2} = 153$.Let $X

Vedclass · Accepted Answer

$\frac{40}{153}$

Vedclass · Answer

(D) Total apples = $3 + 15 = 18$.We draw $2$ apples. The total number of ways to choose $2$ apples is $^{18}C_2 = \frac{18 	imes 17}{2} = 153$.Let $X$ be the number of rotten apples. $X$ can take values $0, 1, 2$.$P(X=0) = \frac{^{15}C_2}{^{18}C_2} = \frac{105}{153}$.$P(X=1) = \frac{^{3}C_1 	imes ^{15}C_1}{^{18}C_2} = \frac{3 	imes 15}{153} = \frac{45}{153}$.$P(X=2) = \frac{^{3}C_2}{^{18}C_2} = \frac{3}{153}$.$E(X) = \sum x_i P(x_i) = 0 	imes \frac{105}{153} + 1 	imes \frac{45}{153} + 2 	imes \frac{3}{153} = \frac{51}{153} = \frac{1}{3}$.$E(X^2) = \sum x_i^2 P(x_i) = 0^2 	imes \frac{105}{153} + 1^2 	imes \frac{45}{153} + 2^2 	imes \frac{3}{153} = \frac{45 + 12}{153} = \frac{57}{153}$.$	ext{Var}(X) = E(X^2) - (E(X))^2 = \frac{57}{153} - (\frac{1}{3})^2 = \frac{57}{153} - \frac{1}{9} = \frac{57}{153} - \frac{17}{153} = \frac{40}{153}$.

Outcome	Probability
$\omega_{1}$	$1/8$
$\omega_{2}$	$2/3$
$\omega_{3}$	$1/3$
$\omega_{4}$	$1/3$
$\omega_{5}$	$-1/4$
$\omega_{6}$	$-1/3$

$x$	$0$	$1$	$2$	$3$
$P(x)$	$0.2$	$k$	$k$	$2k$

$X$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$2k$	$k$	$2k$	$4k$	$k$

Three rotten apples are accidentally mixed with fifteen good apples. Assuming the random variable $X$ to be the number of rotten apples in a draw of two apples,the variance of $X$ is

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