Three balls are drawn at random from a bag co | vedclass

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(C) Total balls = $5 + 4 = 9$. We draw $3$ balls. The total number of ways to draw $3$ balls is $^9C_3 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(C) Total balls = $5 + 4 = 9$. We draw $3$ balls. The total number of ways to draw $3$ balls is $^9C_3 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$. Let $X$ be the number of blue balls. The mean $\bar{X} = E[X] = n 	imes p$,where $n=3$ and $p$ is the probability of drawing a blue ball,$p = \frac{5}{9}$. Thus,$\bar{X} = 3 	imes \frac{5}{9} = \frac{15}{9} = \frac{5}{3}$. Let $Y$ be the number of yellow balls. The mean $\bar{Y} = E[Y] = n 	imes p'$,where $n=3$ and $p'$ is the probability of drawing a yellow ball,$p' = \frac{4}{9}$. Thus,$\bar{Y} = 3 	imes \frac{4}{9} = \frac{12}{9} = \frac{4}{3}$. We need to find $7 \bar{X} + 4 \bar{Y}$: $7 \bar{X} + 4 \bar{Y} = 7 \left(\frac{5}{3}ight) + 4 \left(\frac{4}{3}ight) = \frac{35}{3} + \frac{16}{3} = \frac{51}{3} = 17$.

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$0.10$	$0.12$	$0.20$	$0.08$	$0.07$	$0.05$

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

Three balls are drawn at random from a bag containing $5$ blue and $4$ yellow balls. Let the random variables $X$ and $Y$ respectively denote the number of blue and yellow balls. If $\bar{X}$ and $\bar{Y}$ are the means of $X$ and $Y$ respectively,then $7 \bar{X} + 4 \bar{Y}$ is equal to ..........

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