A bag contains 4 white and 6 black balls. Thr | vedclass

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(C) The total number of balls is $4 + 6 = 10$. Three balls are drawn at random. The total number of ways to draw $3$ balls is $C(10, 3) = \frac{10 	i

Vedclass · Accepted Answer

$56$

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(C) The total number of balls is $4 + 6 = 10$. Three balls are drawn at random. The total number of ways to draw $3$ balls is $C(10, 3) = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$.Let $X$ be the number of white balls. $X$ can take values $0, 1, 2, 3$.$P(X=0) = \frac{C(4,0) 	imes C(6,3)}{120} = \frac{1 	imes 20}{120} = \frac{20}{120} = \frac{1}{6}$.$P(X=1) = \frac{C(4,1) 	imes C(6,2)}{120} = \frac{4 	imes 15}{120} = \frac{60}{120} = \frac{1}{2}$.$P(X=2) = \frac{C(4,2) 	imes C(6,1)}{120} = \frac{6 	imes 6}{120} = \frac{36}{120} = \frac{3}{10}$.$P(X=3) = \frac{C(4,3) 	imes C(6,0)}{120} = \frac{4 	imes 1}{120} = \frac{4}{120} = \frac{1}{30}$.Mean $E(X) = \sum x P(x) = 0(\frac{1}{6}) + 1(\frac{1}{2}) + 2(\frac{3}{10}) + 3(\frac{1}{30}) = 0 + 0.5 + 0.6 + 0.1 = 1.2$.$E(X^2) = \sum x^2 P(x) = 0^2(\frac{1}{6}) + 1^2(\frac{1}{2}) + 2^2(\frac{3}{10}) + 3^2(\frac{1}{30}) = 0 + 0.5 + 1.2 + 0.3 = 2.0$.Variance $\sigma^2 = E(X^2) - [E(X)]^2 = 2.0 - (1.2)^2 = 2.0 - 1.44 = 0.56$.Therefore,$100 \sigma^2 = 100 	imes 0.56 = 56$.

$X = x_{i}$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x_{i})$	$0.1$	$k$	$0.2$	$2k$	$3k$	$k$

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$k$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

$A$ bag contains $4$ white and $6$ black balls. Three balls are drawn at random from the bag. Let $X$ be the number of white balls among the drawn balls. If $\sigma^{2}$ is the variance of $X$,then $100 \sigma^{2}$ is equal to.

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