An urn contains 3 black and 5 red balls. If 3 | vedclass

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(B) Let $X$ be the number of red balls drawn. The total number of balls is $3 + 5 = 8$. We draw $3$ balls from $8$,so the total number of ways is ${}^

Vedclass · Accepted Answer

$\frac{15}{8}$

Vedclass · Answer

(B) Let $X$ be the number of red balls drawn. The total number of balls is $3 + 5 = 8$. We draw $3$ balls from $8$,so the total number of ways is ${}^8 C_3 = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = 56$.The random variable $X$ can take values $0, 1, 2, 3$.$P(X=0) = \frac{{}^5 C_0 	imes {}^3 C_3}{56} = \frac{1 	imes 1}{56} = \frac{1}{56}$$P(X=1) = \frac{{}^5 C_1 	imes {}^3 C_2}{56} = \frac{5 	imes 3}{56} = \frac{15}{56}$$P(X=2) = \frac{{}^5 C_2 	imes {}^3 C_1}{56} = \frac{10 	imes 3}{56} = \frac{30}{56}$$P(X=3) = \frac{{}^5 C_3 	imes {}^3 C_0}{56} = \frac{10 	imes 1}{56} = \frac{10}{56}$The mean $E(X) = \sum x_i P(x_i) = 0 	imes \frac{1}{56} + 1 	imes \frac{15}{56} + 2 	imes \frac{30}{56} + 3 	imes \frac{10}{56} = \frac{0 + 15 + 60 + 30}{56} = \frac{105}{56} = \frac{15}{8}$.

$X = x$	$0$	$1$	$2$
$P(X = x)$	$4k - 10k^2$	$5k - 1$	$3k^3$

An urn contains $3$ black and $5$ red balls. If $3$ balls are drawn at random from the urn,the mean of the probability distribution of the number of red balls drawn is

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