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 $X$ and $Y$ respectively, then $7\bar{X} + 4\bar{Y}$ is equal to _____.

Answer 1

Correct Answer: 17

The total number of ways to select 3 balls from 9 is:
\[ \binom{9}{3} = 84. \]
The probabilities for \(X = r\) (number of blue balls drawn) are:
\[ P(X = r) = \frac{\binom{5}{r} \cdot \binom{4}{3-r}}{\binom{9}{3}}. \]
For \(r = 0\):
\[ P(X = 0) = \frac{\binom{5}{0} \cdot \binom{4}{3}}{84} = \frac{1 \cdot 4}{84} = \frac{4}{84}. \]
For \(r = 1\):
\[ P(X = 1) = \frac{\binom{5}{1} \cdot \binom{4}{2}}{84} = \frac{5 \cdot 6}{84} = \frac{30}{84}. \]
For \(r = 2\):
\[ P(X = 2) = \frac{\binom{5}{2} \cdot \binom{4}{1}}{84} = \frac{10 \cdot 4}{84} = \frac{40}{84}. \]
For \(r = 3\):
\[ P(X = 3) = \frac{\binom{5}{3} \cdot \binom{4}{0}}{84} = \frac{10 \cdot 1}{84} = \frac{10}{84}. \]
The mean of \(X\) is:
\[ \overline{X} = \sum_{r=0}^3 r \cdot P(X = r) = 0 \cdot \frac{4}{84} + 1 \cdot \frac{30}{84} + 2 \cdot \frac{40}{84} + 3 \cdot \frac{10}{84}. \]
\[ \overline{X} = \frac{30 + 80 + 30}{84} = \frac{140}{84} = \frac{5}{3}. \]
Now, compute \(7\overline{X}\):
\[ 7\overline{X} = 7 \cdot \frac{5}{3} = \frac{35}{3}. \]
Similarly, compute probabilities for \(Y = r\) (number of yellow balls drawn):
\[ P(Y = r) = P(X = 3 - r). \]
The mean of \(Y\) is:
\[ \overline{Y} = 3 - \overline{X} = 3 - \frac{5}{3} = \frac{4}{3}. \]
Now, compute \(4\overline{Y}\):
\[ 4\overline{Y} = 4 \cdot \frac{4}{3} = \frac{16}{3}. \]
Finally, compute \(7\overline{X} + 4\overline{Y}\):
\[ 7\overline{X} + 4\overline{Y} = \frac{35}{3} + \frac{16}{3} = \frac{51}{3} = 17. \]
Final Answer: 17.