The problem states that the joint probability mass function (p.m.f.) of $X$ and $Y$ is given by:

$p(x, y) = k(2x + 3y), \quad x = 0, 1, 2; \, y = 1, 2, 3.$

We are tasked with:

1. Finding the marginal distributions of $X$ and $Y$.
2. Finding the conditional probability distributions.
3. Finding the probability distribution of $X + Y$.

Let's go step by step:

1. Finding the constant $k$:

Since the total probability must sum to 1, we can use the following equation:

$\sum_{x=0}^{2} \sum_{y=1}^{3} p(x, y) = 1.$

Substitute $p(x, y) = k(2x + 3y)$ and sum over all possible values of $x$ and $y$:

$\sum_{x=0}^{2} \sum_{y=1}^{3} k(2x + 3y) = 1.$

Now, I'll calculate this to find the value of $k$.The constant $k$ is $\frac{1}{72}$.

Now that we have $k = \frac{1}{72}$, let's move to the next steps.

2. Marginal Distributions

The marginal distribution of $X$ is found by summing the joint probability over all values of $y$:

$p_X(x) = \sum_{y=1}^{3} p(x, y).$

Similarly, the marginal distribution of $Y$ is found by summing the joint probability over all values of $x$:

$p_Y(y) = \sum_{x=0}^{2} p(x, y).$

I'll calculate the marginal distributions for both $X$ and $Y$.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can manually calculate the marginal distributions using the formula:

1. Marginal Distribution of $X$: $p_X(x) = \sum_{y=1}^{3} p(x, y)$ where $p(x, y) = \frac{1}{72} \cdot (2x + 3y)$.

2. Marginal Distribution of $Y$: $p_Y(y) = \sum_{x=0}^{2} p(x, y)$

Let me know if you'd like to continue solving it step by step!

Here are some related questions to explore:

1. How are marginal distributions used in probability theory?
2. What are conditional probability distributions?
3. How can you derive the probability distribution of a sum of two random variables?
4. What are joint probability mass functions used for in real-world applications?
5. How does knowing $p(x, y)$ help in calculating expectations?

Tip:

Always check that the sum of probabilities equals 1 to ensure your probability distribution is valid.