Two bad eggs are mixed accidentally with 10 g | vedclass

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(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 replacemen

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 $Var(X) = E(X^2) - [E(X)]^2$:$Var(X) = \frac{13}{22} - (\frac{1}{2})^2 = \frac{13}{22} - \frac{1}{4} = \frac{26-11}{44} = \frac{15}{44}$

$X = x_i$	$1$	$2$	$3$	$5$
$P(X = x_i)$	$2k^2$	$k$	$k$	$k^2$

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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