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Let's break down the problem and solve it step by step:

### Given:
- \( \mathbb{E}(X) = 1 \)
- \( \mathbb{E}(Y) = 2 \)
- \( \text{Var}(X) = 9 \)
- \( \text{Var}(Y) = 16 \)
- \( \text{Cov}(X, Y) = 6 \)

We are asked to compute:
1. \( \mathbb{E}(2X - Y + 5) \)
2. \( \text{Var}(2X - Y + 5) \)
3. \( \text{Var}(2X - Y + 5) \) when \( X \) and \( Y \) are independent.

### Step 1: Compute \( \mathbb{E}(2X - Y + 5) \)

Using the linearity of expectation:
\[
\mathbb{E}(2X - Y + 5) = 2\mathbb{E}(X) - \mathbb{E}(Y) + 5
\]
Substitute the given values for \( \mathbb{E}(X) = 1 \) and \( \mathbb{E}(Y) = 2 \):
\[
\mathbb{E}(2X - Y + 5) = 2(1) - 2 + 5 = 2 - 2 + 5 = 5
\]

### Step 2: Compute \( \text{Var}(2X - Y + 5) \)

To compute the variance, we use the property of variance for linear combinations:
\[
\text{Var}(aX + bY + c) = a^2 \text{Var}(X) + b^2 \text{Var}(Y) + 2ab \text{Cov}(X, Y)
\]
In our case:
- \( a = 2 \)
- \( b = -1 \)
- \( c = 5 \) (constants do not affect variance)

So:
\[
\text{Var}(2X - Y + 5) = 2^2 \text{Var}(X) + (-1)^2 \text{Var}(Y) + 2(2)(-1) \text{Cov}(X, Y)
\]
Substitute the given values for \( \text{Var}(X) = 9 \), \( \text{Var}(Y) = 16 \), and \( \text{Cov}(X, Y) = 6 \):
\[
\text{Var}(2X - Y + 5) = 4(9) + 1(16) + 2(2)(-1)(6)
\]
\[
\text{Var}(2X - Y + 5) = 36 + 16 - 24 = 28
\]

### Step 3: Compute \( \text{Var}(2X - Y + 5) \) when \( X \) and \( Y \) are independent

When \( X \) and \( Y \) are independent, the covariance \( \text{Cov}(X, Y) = 0 \). Thus, the formula for variance becomes:
\[
\text{Var}(2X - Y + 5) = 2^2 \text{Var}(X) + (-1)^2 \text{Var}(Y)
\]
Substitute the values for \( \text{Var}(X) \) and \( \text{Var}(Y) \):
\[
\text{Var}(2X - Y + 5) = 4(9) + 1(16)
\]
\[
\text{Var}(2X - Y + 5) = 36 + 16 = 52
\]

### Final Answers:
1. \( \mathbb{E}(2X - Y + 5) = 5 \)
2. \( \text{Var}(2X - Y + 5) = 28 \)
3. \( \text{Var}(2X - Y + 5) \) when \( X \) and \( Y \) are independent = 52

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Would you like a deeper explanation on any step?

Here are 5 related questions to further explore:
1. How does the covariance affect the variance of a linear combination of two random variables?
2. What is the significance of the independence of random variables in calculating variance?
3. Can you generalize the rule for the variance of a sum of random variables?
4. How would the calculation change if \( X \) and \( Y \) were negatively correlated?
5. How is the expectation of a linear combination of random variables derived?

**Tip:** Always check the independence of random variables, as it simplifies variance calculations by eliminating covariance terms.

Suppose that the random variables X and Y have: E(X) = 1, E(Y) = 2, Var(X) = 9, Var(Y) = 16, Cov(X,Y) = 6. Compute E(2X - Y + 5) and Var(2X - Y + 5). What is Var(2X - Y + 5) when X and Y are independent?

Learn how to compute the expectation and variance of the linear combination 2X - Y + 5 using properties of expectation and variance, including the role of covariance when X and Y are dependent. Understand how independence simplifies variance calculations.

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