JEE MAIN - Mathematics (2024 - 8th April Morning Shift - No. 26)
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 ___________.
Answer
17
Explanation
| $$X$$ | 3 | 2 | 1 | 0 |
|---|---|---|---|---|
| $$Y$$ | 0 | 1 | 2 | 3 |
$$\begin{aligned} & \bar{X}=\sum X p(X) \\ & \bar{Y}=\sum Y p(Y) \\ & P(X=3)=P(Y=0)=\frac{{ }^5 C_3 \cdot C_0}{{ }^9 C_3}=\frac{{ }^5 C_2}{{ }^9 C_3}=\frac{5}{42} \\ & P(X=2)=P(Y=1)=\frac{{ }^5 C_2 \cdot C_1}{{ }^9 C_3}=\frac{10}{21} \\ & P(X=1)=P(Y=2)=\frac{{ }^5 C_1 \cdot C_2}{{ }^9 C_3}=\frac{5}{14} \\ & P(X=0)=P(Y=3)=\frac{{ }^5 C_0 \cdot C_3}{{ }^9 C_3}=\frac{4}{84}=\frac{1}{21} \\ & \bar{X}=3 \times \frac{5}{42}+2 \times \frac{10}{21}+\frac{5}{14}+0 \times \frac{1}{21}=\frac{15+40+15}{42}=\frac{70}{42} \\ & \bar{Y}=0 \times \frac{5}{42}+1 \times \frac{10}{21}+2 \times \frac{5}{14}+3 \times \frac{1}{21}=\frac{20+30+6}{42}=\frac{56}{42} \\ & 7 \bar{X}+4 \bar{Y}=17 \end{aligned}$$
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