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

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.