Two bad eggs are mixed accidentally with 10 g | vedclass

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Question

(B) Let $X$ be the random variable representing the number of bad eggs drawn. The total number of eggs is $12$, and we draw $3$ eggs without replaceme

Vedclass · Accepted Answer

$\frac{15}{44}$

Vedclass · Answer

(B) Let $X$ be the random variable representing the number of bad eggs drawn. The total number of eggs is $12$, and we draw $3$ eggs without replacement. The possible values for $X$ are $0, 1, 2$.The probability distribution is calculated as follows:$P(X=0) = \frac{{}^{10}C_3}{{}^{12}C_3} = \frac{120}{220} = \frac{12}{22}$$P(X=1) = \frac{{}^{10}C_2 	imes {}^{2}C_1}{{}^{12}C_3} = \frac{45 	imes 2}{220} = \frac{90}{220} = \frac{9}{22}$$P(X=2) = \frac{{}^{10}C_1 	imes {}^{2}C_2}{{}^{12}C_3} = \frac{10 	imes 1}{220} = \frac{10}{220} = \frac{1}{22}$Now, we calculate the mean $E(X) = \sum x_i P_i$:$E(X) = (0 	imes \frac{12}{22}) + (1 	imes \frac{9}{22}) + (2 	imes \frac{1}{22}) = \frac{9+2}{22} = \frac{11}{22} = \frac{1}{2}$Next, we calculate $E(X^2) = \sum x_i^2 P_i$:$E(X^2) = (0^2 	imes \frac{12}{22}) + (1^2 	imes \frac{9}{22}) + (2^2 	imes \frac{1}{22}) = \frac{9+4}{22} = \frac{13}{22}$Finally, the variance is given by $Var(X) = E(X^2) - [E(X)]^2$:$Var(X) = \frac{13}{22} - (\frac{1}{2})^2 = \frac{13}{22} - \frac{1}{4} = \frac{26-11}{44} = \frac{15}{44}$

$X = x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X = x)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2 + k$

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$K + \frac{2}{10}$	$K + \frac{3}{10}$	$K + \frac{3}{10}$	$K + \frac{4}{10}$	$K + \frac{2}{10}$

$X = x_i$	$3$	$5$	$7$	$9$
$P(X = x_i)$	$k$	$2k$	$3k$	$4k$

Two bad eggs are mixed accidentally with $10$ good ones. If three eggs are drawn at random from this lot in succession without replacement, then the variance of the probability distribution of the number of bad eggs drawn is

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