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

Question

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 $X$

Accepted Answer

$X$
 3
 2
 1
 0

$Y$
 0
 1
 2
 3

$\begin{gathered} \vec{X}=\sum Xp(X) \ \vec{Y}=\sum Yp(Y) \ P(X=3)=P(Y=0)=\frac{{}^{5}{C}_{3}\cdot {C}_{0}}{{}^{9}{C}_{3}}=\frac{{}^{5}{C}_{2}}{{}^{9}{C}_{3}}=\frac{5}{42} \ P(X=2)=P(Y=1)=\frac{{}^{5}{C}_{2}\cdot {C}_{1}}{{}^{9}{C}_{3}}=\frac{10}{21} \ P(X=1)=P(Y=2)=\frac{{}^{5}{C}_{1}\cdot {C}_{2}}{{}^{9}{C}_{3}}=\frac{5}{14} \ P(X=0)=P(Y=3)=\frac{{}^{5}{C}_{0}\cdot {C}_{3}}{{}^{9}{C}_{3}}=\frac{4}{84}=\frac{1}{21} \ \vec{X}=3	imes \frac{5}{42}+2	imes \frac{10}{21}+\frac{5}{14}+0	imes \frac{1}{21}=\frac{15+40+15}{42}=\frac{70}{42} \ \vec{Y}=0	imes \frac{5}{42}+1	imes \frac{10}{21}+2	imes \frac{5}{14}+3	imes \frac{1}{21}=\frac{20+30+6}{42}=\frac{56}{42} \ 7\vec{X}+4\vec{Y}=17 \end{gathered}$

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