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(C) The total number of balls is $5 + 3 = 8$. Two balls are drawn at random. The total number of ways to draw $2$ balls is ${}^8C_2 = \frac{8 	imes 7

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(C) The total number of balls is $5 + 3 = 8$. Two balls are drawn at random. The total number of ways to draw $2$ balls is ${}^8C_2 = \frac{8 	imes 7}{2} = 28$. Let $X$ be the random variable representing the number of white balls drawn. The possible values for $X$ are $0, 1, 2$. $P(X = 0) = \frac{{}^5C_2}{{}^8C_2} = \frac{10}{28}$. $P(X = 1) = \frac{{}^5C_1 	imes {}^3C_1}{{}^8C_2} = \frac{5 	imes 3}{28} = \frac{15}{28}$. $P(X = 2) = \frac{{}^3C_2}{{}^8C_2} = \frac{3}{28}$. The mean of $X$ is given by $E(X) = \sum x_i P(x_i) = 0 	imes P(X = 0) + 1 	imes P(X = 1) + 2 	imes P(X = 2)$. $E(X) = 0 	imes \frac{10}{28} + 1 	imes \frac{15}{28} + 2 	imes \frac{3}{28} = 0 + \frac{15}{28} + \frac{6}{28} = \frac{21}{28} = \frac{3}{4}$.

$X = x$	$0$	$1$	$2$
$P(X = x)$	$\frac{1}{5}$	$\frac{2}{5}$	$\frac{2}{5}$

Two balls are drawn at random from a bag containing $5$ black balls and $3$ white balls. If the random variable $X$ denotes the number of white balls drawn,then the mean of $X$ is

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