From a bag containing 4 white and 5 red balls | vedclass

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Question

(A) Let $X$ be the random variable representing the number of red balls drawn. The total number of balls is $4 + 5 = 9$. The number of ways to draw $3

Vedclass · Accepted Answer

$\frac{5}{3}$

Vedclass · Answer

(A) Let $X$ be the random variable representing the number of red balls drawn. The total number of balls is $4 + 5 = 9$. The number of ways to draw $3$ balls from $9$ is ${}^9C_3 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$.The probability distribution of $X$ is:$P(X=0) = \frac{{}^4C_3}{{}^9C_3} = \frac{4}{84}$$P(X=1) = \frac{{}^4C_2 	imes {}^5C_1}{{}^9C_3} = \frac{6 	imes 5}{84} = \frac{30}{84}$$P(X=2) = \frac{{}^4C_1 	imes {}^5C_2}{{}^9C_3} = \frac{4 	imes 10}{84} = \frac{40}{84}$$P(X=3) = \frac{{}^4C_0 	imes {}^5C_3}{{}^9C_3} = \frac{1 	imes 10}{84} = \frac{10}{84}$The mean $E(X) = \sum x_i P(x_i) = 0 	imes \frac{4}{84} + 1 	imes \frac{30}{84} + 2 	imes \frac{40}{84} + 3 	imes \frac{10}{84}$$E(X) = \frac{0 + 30 + 80 + 30}{84} = \frac{140}{84} = \frac{5}{3}$.Thus,the mean is $\frac{5}{3}$. Hence,option $(A)$ is correct.

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

From a bag containing $4$ white and $5$ red balls,if $3$ balls are drawn at random,then the mean of the number of red balls among the balls drawn is:

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