Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables X and Y respectively ​

Question

Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables \(\mathrm{X}\) and \(\mathrm{Y}\) 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 ______.

Answer 1

Correct answer is : 17  

\(\mathrm{X}=1 \)

\( \frac{{ }^5 \mathrm{C}_1 \times{ }^4 \mathrm{C}_2}{{ }^9 \mathrm{C}_3}=\frac{5 \times 6}{84}=\frac{5}{14} \)

\(\mathrm{X}=2 \)

\( \frac{{ }^5 \mathrm{C}_2 \times{ }^4 \mathrm{C}_1}{{ }^9 \mathrm{C}_3}=\frac{10 \times 4}{84}=\frac{10}{21} \)

\(\mathrm{X}=3\) 

 \( \frac{{ }^5 \mathrm{C}_3}{{ }^9 \mathrm{C}_3}=\frac{5 \times 4 \times 3}{9 \times 8 \times 7}=\frac{5}{42}\)

\(\mathrm{Y}=1=\frac{10}{21} \)

\(\mathrm{Y}=2=\frac{5}{14}\)

\(\mathrm{Y}=3=\frac{{ }^4 \mathrm{C}_3}{{ }^9 \mathrm{C}_3}=\frac{4 \times 3 \times 2}{9 \times 8 \times 7}=\frac{1}{21}\)

\(\bar{X}=p(x=1) x_1+p(x=2) x_2+p(x=3) x_3\)

\( \bar{X}=\frac{5}{14}+\frac{20}{21}+\frac{15}{42} \)

\( \bar{X}=\frac{30+80+30}{84}=\frac{140}{84}\)

\( \bar{Y}=p(y=1) y_1+p(y=2) y_2+p(y=3) y_3 \)

\(\bar{Y}=\frac{20+30+6}{42}=\frac{20+30+6}{42}=\frac{56}{42}\)

\(=7 \bar{X}+4 \bar{Y}\)

\(=\frac{140}{12}+\frac{56}{21} \times 2=\frac{980+448}{84}=\frac{1428}{84}=17 \text { Ans. }\)