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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Aicharya · Accepted Answer

$\begin{aligned} & 7 \overline{\mathrm{x}}=\frac{{ }^5 \mathrm{C}_1{ }^4 \mathrm{C}_2+{ }^5 \mathrm{C}_2 \cdot{ }^4 \mathrm{C}_1 \times 2+{ }^5 \mathrm{C}_3 \cdot{ }^4 \mathrm{C}_0 \times 3}{{ }^9 \mathrm{C}_3} \times 7 \\ & \frac{30+80+30}{84} \times 7 \\ & =\frac{140}{12}=\frac{70}{6}=\frac{35}{3}\end{aligned}$
$\begin{array}{|l|l|l|l|l|l|}
\hline yellow & 0 & 1 & 2 & 3 & 4 \\
\hline & & { }^5 \mathrm{C}_2{ }^4 \mathrm{C}_1 & { }^5 \mathrm{C}_1{ }^4 \mathrm{C}_2 & { }^5 \mathrm{C}_0{ }^4 \mathrm{C}_3 & 0 \\
\hline
\end{array}$
$4 \bar{y}=\frac{40+60+12}{84} \times 4=\frac{112}{21}=\frac{16}{3}$

Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables…

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