The joint probability distribution $f(x, y) = \frac{x + y}{30}$ is defined for the values of $x$ and $y$ in the set $\{0, 1, 2, 3\}$. We need to calculate the following probabilities:

(i) $P[ X \leq 2 , Y = 1 ]$

This represents the probability that $X \leq 2$ and $Y = 1$. We sum the joint probabilities where $Y = 1$ and $X$ can take values $0, 1, 2$.

[ P[ X \leq 2, Y = 1 ] = P[ X = 0, Y = 1 ] + P[ X = 1, Y = 1 ] + P[ X = 2, Y = 1 ] ]

Using the joint probability function $f(x, y) = \frac{x + y}{30}$, we can compute:

$P[ X = 0, Y = 1 ] = \frac{0 + 1}{30} = \frac{1}{30}$ $P[ X = 1, Y = 1 ] = \frac{1 + 1}{30} = \frac{2}{30}$ $P[ X = 2, Y = 1 ] = \frac{2 + 1}{30} = \frac{3}{30}$

So,

$P[ X \leq 2, Y = 1 ] = \frac{1}{30} + \frac{2}{30} + \frac{3}{30} = \frac{6}{30} = \frac{1}{5}$

(ii) $P[ X > 2, Y \leq 1 ]$

This represents the probability that $X > 2$ and $Y \leq 1$. For $X > 2$, the only value of $X$ that satisfies this condition is $X = 3$. So, we need to sum the joint probabilities where $X = 3$ and $Y$ can take the values $0$ and $1$.

[ P[ X > 2, Y \leq 1 ] = P[ X = 3, Y = 0 ] + P[ X = 3, Y = 1 ] ]

Using the joint probability function $f(x, y) = \frac{x + y}{30}$, we can compute:

$P[ X = 3, Y = 0 ] = \frac{3 + 0}{30} = \frac{3}{30}$ $P[ X = 3, Y = 1 ] = \frac{3 + 1}{30} = \frac{4}{30}$

So,

$P[ X > 2, Y \leq 1 ] = \frac{3}{30} + \frac{4}{30} = \frac{7}{30}$

(iii) $P[ X > Y ]$

This represents the probability that $X > Y$. We need to find the joint probabilities where $X > Y$. These pairs are as follows:

• For $X = 1$, $Y = 0$ satisfies $X > Y$.
• For $X = 2$, $Y = 0, 1$ satisfy $X > Y$.
• For $X = 3$, $Y = 0, 1, 2$ satisfy $X > Y$.

Now, sum the probabilities for these cases:

[ P[ X > Y ] = P[ X = 1, Y = 0 ] + P[ X = 2, Y = 0 ] + P[ X = 2, Y = 1 ] + P[ X = 3, Y = 0 ] + P[ X = 3, Y = 1 ] + P[ X = 3, Y = 2 ] ]

Using the joint probability function $f(x, y) = \frac{x + y}{30}$, we can compute each term:

$P[ X = 1, Y = 0 ] = \frac{1 + 0}{30} = \frac{1}{30}$ $P[ X = 2, Y = 0 ] = \frac{2 + 0}{30} = \frac{2}{30}$ $P[ X = 2, Y = 1 ] = \frac{2 + 1}{30} = \frac{3}{30}$ $P[ X = 3, Y = 0 ] = \frac{3 + 0}{30} = \frac{3}{30}$ $P[ X = 3, Y = 1 ] = \frac{3 + 1}{30} = \frac{4}{30}$ $P[ X = 3, Y = 2 ] = \frac{3 + 2}{30} = \frac{5}{30}$

Now, sum them up:

$P[ X > Y ] = \frac{1}{30} + \frac{2}{30} + \frac{3}{30} + \frac{3}{30} + \frac{4}{30} + \frac{5}{30} = \frac{18}{30} = \frac{3}{5}$

Final Answers:

1. $P[ X \leq 2 , Y = 1 ] = \frac{1}{5}$
2. $P[ X > 2, Y \leq 1 ] = \frac{7}{30}$
3. $P[ X > Y ] = \frac{3}{5}$

Would you like further details on any of the calculations, or have any questions about the concepts used?

Here are some related questions for further practice:

1. What is the marginal probability $P(X = 1)$?
2. How do you calculate the expected value $E[X]$ for this distribution?
3. Find the conditional probability $P(Y = 0 \mid X = 3)$.
4. Compute the variance of $X$ in this distribution.
5. What is the covariance between $X$ and $Y$?

Tip: Always check the sum of all joint probabilities in a distribution; it should equal 1 to ensure it is a valid probability distribution.