To find the conditional probability mass function (PMF) $f(y \mid x = 1)$, we use the formula for conditional probability:

$f(y \mid x = 1) = \frac{f(x = 1, y)}{P(X = 1)}$

Step 1: Find $P(X = 1)$

To find $P(X = 1)$, we sum the joint probabilities for all values of $y$ when $X = 1$:

$P(X = 1) = f(1, 0) + f(1, 1)$

From the given joint PMF:

• $f(1, 0) = \frac{1}{2}$
• $f(1, 1) = \frac{1}{8}$

Thus,

$P(X = 1) = \frac{1}{2} + \frac{1}{8} = \frac{4}{8} + \frac{1}{8} = \frac{5}{8}$

Step 2: Compute the conditional PMF $f(y \mid x = 1)$

Now, we calculate $f(y \mid x = 1)$ for each possible value of $y$ (which can be 0 or 1):

• For $y = 0$:

$f(0 \mid x = 1) = \frac{f(1, 0)}{P(X = 1)} = \frac{\frac{1}{2}}{\frac{5}{8}} = \frac{1}{2} \times \frac{8}{5} = \frac{4}{5}$

• For $y = 1$:

$f(1 \mid x = 1) = \frac{f(1, 1)}{P(X = 1)} = \frac{\frac{1}{8}}{\frac{5}{8}} = \frac{1}{8} \times \frac{8}{5} = \frac{1}{5}$

Final Conditional PMF

Thus, the conditional PMF $f(y \mid x = 1)$ is:

\begin{cases} \frac{4}{5} & \text{if } y = 0 \\ \frac{1}{5} & \text{if } y = 1 \end{cases}$$ Let me know if you'd like further details or have any questions! Here's a list of related questions for further exploration: 1. How do you derive the marginal PMF from a joint PMF? 2. What is the definition of conditional independence in probability? 3. How do you compute the expected value of a discrete random variable? 4. What is the relationship between joint and conditional distributions in probability theory? 5. Can you explain Bayes' theorem and its applications in probability? **Tip**: Always remember that the sum of probabilities in a conditional distribution must equal 1. For example, $$f(0 \mid x = 1) + f(1 \mid x = 1) = 1$$, which you can use as a quick check.