Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables and respectively denote the number of blue and yellow balls. If and are the means of and respectively, then is equal to

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Solution

Q.
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 ________ . [2024]

Ans.
(17)

Blue balls (X) 0 1 2 3

Probability P(X) $\frac{{}^{5}{C_{0} {}^{4}{C_{3}}}}{{}^{9}{C_{3}}}$ $\frac{{}^{5}C_{1} \times {}^{4}{C_{2}}}{{}^{9}{C_{3}}}$ $\frac{{}^{5}{C_{2} \times {}^{4}{C_{1}}}}{{}^{9}{C_{3}}}$ $\frac{{}^{5}{C_{3} \times {}^{4}{C_{0}}}}{{}^{9}{C_{3}}}$

$\bar{X} = \frac{0 + 5 \times 6 + 2 \times 10 \times 4 + 3 \times 10 \times 1}{84}$

$\Rightarrow \bar{X} = \frac{140}{84} \Rightarrow 7 \bar{X} = \frac{140}{12} = \frac{35}{3}$

Yellow balls (X) 0 1 2 3

Probability P(Y) $\frac{{}^{5}{C_{3} \times {}^{4}{C_{0}}}}{{}^{9}{C_{3}}}$ $\frac{{}^{5}{C_{2} \times {}^{4}{C_{1}}}}{{}^{9}{C_{3}}}$ $\frac{{}^{5}{C_{1} \times {}^{4}{C_{2}}}}{{}^{9}{C_{3}}}$ $\frac{{}^{5}{C_{0} \times {}^{4}{C_{3}}}}{{}^{9}{C_{3}}}$

$\bar{Y} = \frac{0 + 10 \times 4 + 2 \times 5 \times 6 + 3 \times 4}{84} = \frac{112}{84} \Rightarrow 4 \bar{Y} = \frac{16}{3}$

$7 \bar{X} + 4 \bar{Y} = \frac{35}{3} + \frac{16}{3} = 17$

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