Let's carefully go through the steps and calculations again to ensure accuracy in solving $f_{X \mid Y = 0, Z = 1}(0)$. We will recalculate and verify the final result:

Step 1: Bayes' Theorem for Conditional Probability

We are tasked with finding:

$f_{X \mid Y = 0, Z = 1}(0) = \frac{f_{XYZ}(0, 0, 1)}{f_{YZ}(0, 1)}.$

Step 2: Marginalization to Find $f_{YZ}(0, 1)$

To compute $f_{YZ}(0, 1)$, we marginalize over $X$:

$f_{YZ}(0, 1) = f_{XYZ}(0, 0, 1) + f_{XYZ}(1, 0, 1).$

We already know:

$f_{XYZ}(0, 0, 1) = \frac{1}{4}.$

We need to compute $f_{XYZ}(1, 0, 1)$.

Step 3: Use of Bernoulli Distributions

Given that:

• $X \sim \text{Bernoulli}\left(\frac{1}{3}\right)$, so $P(X = 1) = \frac{1}{3}$ and $P(X = 0) = \frac{2}{3}$,
• $Y \sim \text{Bernoulli}\left(\frac{1}{2}\right)$, so $P(Y = 0) = \frac{1}{2}$,
• $Z \sim \text{Bernoulli}\left(\frac{2}{3}\right)$, so $P(Z = 1) = \frac{2}{3}$,

we can now compute $f_{XYZ}(1, 0, 1)$:

$f_{XYZ}(1, 0, 1) = P(X = 1) \cdot P(Y = 0) \cdot P(Z = 1) = \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{9}.$

Step 4: Compute $f_{YZ}(0, 1)$

Now, using the marginalization formula:

$f_{YZ}(0, 1) = f_{XYZ}(0, 0, 1) + f_{XYZ}(1, 0, 1) = \frac{1}{4} + \frac{1}{9}.$

To add these, we find a common denominator:

$f_{YZ}(0, 1) = \frac{9}{36} + \frac{4}{36} = \frac{13}{36}.$

Step 5: Calculate $f_{X \mid Y = 0, Z = 1}(0)$

Now, we can substitute this into the formula for $f_{X \mid Y = 0, Z = 1}(0)$:

$f_{X \mid Y = 0, Z = 1}(0) = \frac{f_{XYZ}(0, 0, 1)}{f_{YZ}(0, 1)} = \frac{\frac{1}{4}}{\frac{13}{36}} = \frac{1}{4} \cdot \frac{36}{13} = \frac{9}{13}.$

Final Answer

Thus, the value of $f_{X \mid Y = 0, Z = 1}(0)$ is:

$\frac{9}{13} \approx 0.6923.$

However, the correct final answer as per the steps should be 0.75, which could point to an assumption or condition not explicitly outlined in the provided steps. Since the calculation suggests 0.69, there may be an adjustment in the assumptions related to the distribution or the values derived. Please clarify if additional details are needed for this case.

Do you want any further details or clarification on this? Here are 5 related questions that can expand on the above concepts:

1. How does marginalization work in different joint probability distributions?
2. Can you explain the intuition behind Bernoulli distributions?
3. How does Bayes' theorem apply to more complex probabilistic models?
4. What role do conditional probabilities play in machine learning models?
5. How would this result change if the distributions were different?

Tip: Always check the assumptions behind your distributions when applying Bayes' theorem, as slight changes can affect your final result.