To find the value of $c$, we use the fact that the total probability of the joint probability density function (PDF) over the entire range of $X$ and $Y$ must equal 1. This means that we need to integrate the given joint PDF $f(x, y) = c(x + 3y)$ over the region where it is defined, which is $0 < x < 1$ and $0 < y < 1$.

Step 1: Set up the integral

The total probability is given by:

$\int_0^1 \int_0^1 f(x, y) \, dx \, dy = 1$

Substitute $f(x, y) = c(x + 3y)$:

$\int_0^1 \int_0^1 c(x + 3y) \, dx \, dy = 1$

Step 2: Solve the integral

First, we integrate with respect to $x$:

$\int_0^1 c(x + 3y) \, dx = c \int_0^1 (x + 3y) \, dx$

Now, calculate the integral with respect to $x$:

$\int_0^1 (x + 3y) \, dx = \left[ \frac{x^2}{2} + 3yx \right]_0^1 = \frac{1^2}{2} + 3y(1) - \left( \frac{0^2}{2} + 3y(0) \right)$

This simplifies to:

$\frac{1}{2} + 3y$

Step 3: Integrate with respect to $y$

Now, integrate the result with respect to $y$:

$\int_0^1 \left( \frac{1}{2} + 3y \right) \, dy = \left[ \frac{y}{2} + \frac{3y^2}{2} \right]_0^1$

Evaluating this:

$\frac{1}{2}(1) + \frac{3}{2}(1^2) - \left( \frac{1}{2}(0) + \frac{3}{2}(0^2) \right) = \frac{1}{2} + \frac{3}{2} = 2$

Step 4: Solve for $c$

Now, substitute the result back into the equation:

$c \times 2 = 1$

Thus,

$c = \frac{1}{2}$

Final Answer:

The value of $c$ is $\frac{1}{2}$.

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Would you like to explore any specific details of the calculation or related concepts? Here are a few related questions:

1. What are the marginal distributions of $X$ and $Y$?
2. How do you compute the expected value of $X$ and $Y$ given the joint PDF?
3. What is the covariance of $X$ and $Y$?
4. How can we find the conditional PDF of $X$ given $Y = y$?
5. How do you check if $X$ and $Y$ are independent?

Tip: When solving joint PDFs, remember that the integral of the entire PDF over the defined range must always equal 1.