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Let's solve each part of the question:

### Given Information
We have two independent random variables \( X \) and \( Y \) with their joint probability distribution \( f(x, y) \) shown in the table. Specifically:

- Values for \( X \): 2 and 4
- Values for \( Y \): 1, 3, and 5

The joint probabilities are given as follows:
\[
\begin{array}{c|cc}
f(x, y) & X=2 & X=4 \\
\hline
Y=1 & 0.10 & 0.15 \\
Y=3 & 0.20 & 0.30 \\
Y=5 & 0.10 & 0.15 \\
\end{array}
\]

#### Part (a): \( E(2X - 3Y) \)
The expected value \( E(2X - 3Y) \) can be computed by breaking it down into:
\[
E(2X - 3Y) = 2E(X) - 3E(Y)
\]
Since \( X \) and \( Y \) are independent, we can find \( E(X) \) and \( E(Y) \) separately using their marginal distributions.

1. **Calculate \( E(X) \):**
   \[
   E(X) = \sum_{x} x \cdot P(X=x)
   \]
   where
   - \( P(X=2) = 0.10 + 0.20 + 0.10 = 0.40 \)
   - \( P(X=4) = 0.15 + 0.30 + 0.15 = 0.60 \)

   Then,
   \[
   E(X) = 2 \cdot 0.40 + 4 \cdot 0.60 = 0.80 + 2.4 = 3.2
   \]

2. **Calculate \( E(Y) \):**
   \[
   E(Y) = \sum_{y} y \cdot P(Y=y)
   \]
   where
   - \( P(Y=1) = 0.10 + 0.15 = 0.25 \)
   - \( P(Y=3) = 0.20 + 0.30 = 0.50 \)
   - \( P(Y=5) = 0.10 + 0.15 = 0.25 \)

   Then,
   \[
   E(Y) = 1 \cdot 0.25 + 3 \cdot 0.50 + 5 \cdot 0.25 = 0.25 + 1.5 + 1.25 = 3.0
   \]

3. **Calculate \( E(2X - 3Y) \):**
   \[
   E(2X - 3Y) = 2 \cdot 3.2 - 3 \cdot 3.0 = 6.4 - 9 = -2.6
   \]

#### Part (b): \( E(XY) \)
Since \( X \) and \( Y \) are independent, \( E(XY) = E(X) \cdot E(Y) \).

Using the values calculated above:
\[
E(XY) = E(X) \cdot E(Y) = 3.2 \cdot 3.0 = 9.6
\]

### Final Answers
1. \( E(2X - 3Y) = -2.6 \)
2. \( E(XY) = 9.6 \)

Would you like further explanation or details on any step?

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Here are some related questions that may deepen your understanding:

1. How would you find the variance of \( X \) and \( Y \) using the marginal distributions?
2. What is the significance of \( X \) and \( Y \) being independent in this problem?
3. How would the calculation change if \( X \) and \( Y \) were not independent?
4. Can you calculate \( \text{Var}(2X - 3Y) \) using the variances of \( X \) and \( Y \)?
5. How would we approach this problem if it were given as a continuous joint distribution?

**Tip:** When variables are independent, \( E(XY) = E(X) \cdot E(Y) \) simplifies the calculation significantly, as we can treat their expectations separately.

Suppose that X and Y are independent random variables having the joint probability distribution shown. Find (a) E(2X - 3Y); (b) E(XY).

Learn how to calculate the expected values E(2X - 3Y) and E(XY) for independent random variables X and Y using joint probability distributions and properties of expectations. Suitable for college-level probability courses.

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