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 solve the problem, we need to compute the expected values (means) of the random variables $X$ and $Y$, which represent the number of blue and yellow balls drawn from the bag, respectively. We have a total of 9 balls: 5 blue and 4 yellow.

Step 1: Compute the Mean of $X$

$X$ denotes the number of blue balls drawn. The expectation $E(X)$, or $\bar{X}$, can be calculated using the hypergeometric distribution formula:

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

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

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

Step 2: Compute the Mean of $Y$

$Y$ denotes the number of yellow balls drawn. Using a similar approach, we find:

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

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

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

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

We have $\bar{X} = \frac{5}{3}$ and $\bar{Y} = \frac{4}{3}$. Plug these into the required 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 calculated value is 17, which matches the given expected range (17, 17). Hence, our computed solution is correct.

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

Answer 2

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.