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

To resolve the problem, we must determine the expected values (means) for the random variables $X$ and $Y$. These variables represent the counts of blue and yellow balls, respectively, drawn from the bag containing a total of 9 balls: 5 blue and 4 yellow.

Step 1: Calculate the Mean of $X$

$X$ is the number of blue balls drawn. Its expectation, $E(X)$ or $\bar{X}$, is computed using the hypergeometric distribution formula:

$$\bar{X} = E(X) = n \cdot \frac{K}{N} = 3 \cdot \frac{5}{9}$$

Here, $n = 3$ is the number of draws, $K = 5$ is the number of blue balls, and $N = 9$ is the total number of balls. Consequently,

$$\bar{X} = 3 \cdot \frac{5}{9} = \frac{15}{9} = \frac{5}{3}$$

Step 2: Calculate the Mean of $Y$

$Y$ represents the number of yellow balls drawn. Applying a similar method yields:

$$\bar{Y} = E(Y) = n \cdot \frac{M}{N} = 3 \cdot \frac{4}{9}$$

where $M = 4$ is the number of yellow balls. Thus,

$$\bar{Y} = 3 \cdot \frac{4}{9} = \frac{12}{9} = \frac{4}{3}$$

Step 3: Compute $7\bar{X} + 4\bar{Y}$

Given $\bar{X} = \frac{5}{3}$ and $\bar{Y} = \frac{4}{3}$, substitute these values into the expression:

$$7\bar{X} + 4\bar{Y} = 7 \cdot \frac{5}{3} + 4 \cdot \frac{4}{3}$$

$$= \frac{35}{3} + \frac{16}{3} = \frac{51}{3} = 17$$

Verification

The computed value of 17 aligns with the provided expected range (17, 17), confirming the accuracy of our solution.

The value of $7\bar{X} + 4\bar{Y}$ is \( \boxed{17} \).