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 ________

We consider drawing 3 balls from a bag containing 5 blue and 4 yellow balls. Let X and Y be the number of blue and yellow balls drawn, respectively, and we seek to evaluate 7$$\bar{X}$$ + 4$$\bar{Y}$$.

Since the total number of balls is 9 and we draw 3 without replacement, by the hypergeometric distribution we have $$\bar{X}=E[X]=3\times\frac{5}{9}=\frac{15}{9}=\frac{5}{3}$$ and $$\bar{Y}=E[Y]=3\times\frac{4}{9}=\frac{12}{9}=\frac{4}{3}$$. Notice that X+Y=3 so $$\bar{X}+\bar{Y}=3$$, which checks as $$\tfrac{5}{3}+\tfrac{4}{3}=3$$.

Substituting these into the expression gives $$7\bar{X}+4\bar{Y}=7\times\frac{5}{3}+4\times\frac{4}{3}=\frac{35}{3}+\frac{16}{3}=\frac{51}{3}=17$$.

Therefore, the final answer is 17.