To solve this problem, we need to determine $f_{Y|X=x}(1)$, which is the conditional probability $P(Y = 1 \mid X = 3)$. Let's go through the process step by step.

Step 1: Understand the table and the question

The table provides the joint probability mass function (pmf) of two random variables $X$ and $Y$. The values in the table represent $P(X = x, Y = y)$ for different values of $x$ and $y$. We are given the marginal probabilities $f_X(x)$ and $f_Y(y)$ as well.

Step 2: Extract the relevant joint probabilities

We need to find $P(Y = 1 \mid X = 3)$. This is given by: $P(Y = 1 \mid X = 3) = \frac{P(X = 3, Y = 1)}{P(X = 3)}$

Step 3: Find $P(X = 3, Y = 1)$ from the table

From the table: $P(X = 3, Y = 1) = a_{31}$

Step 4: Find $P(X = 3)$

$P(X = 3)$ is given directly as $f_X(3) = 0.45$.

Step 5: Determine $P(Y = 1 \mid X = 3)$

To solve for $P(Y = 1 \mid X = 3)$, we need to find the value of $a_{31}$. The table provides the marginal probability $f_Y(1) = 0.3$. This marginal probability is the sum of all joint probabilities where $Y = 1$: $f_Y(1) = P(X = 1, Y = 1) + P(X = 2, Y = 1) + P(X = 3, Y = 1)$ Using the table, we have: $f_Y(1) = 0.05 + a_{21} + a_{31} = 0.3$ Given that $f_Y(1) = 0.3$, and knowing $P(X = 1, Y = 1) = 0.05$, we can solve for $a_{21} + a_{31}$: $a_{21} + a_{31} = 0.3 - 0.05 = 0.25$

Now, to find $a_{31}$, we subtract $a_{21}$ from this result. But $a_{21}$ is unknown, so we would need additional information to solve this. However, if we were only given that $a_{31}$ is the remaining part, then: $a_{31} = 0.25 - a_{21}$

Given the values: $P(Y = 1 \mid X = 3) = \frac{a_{31}}{0.45}$ Without knowing $a_{21}$, we cannot definitively solve for $P(Y = 1 \mid X = 3)$ but we can express it as: $P(Y = 1 \mid X = 3) = \frac{0.25 - a_{21}}{0.45}$

Final Step: In case $a_{21}$ was provided or deduced

If $a_{21}$ were provided, you would simply substitute and find $P(Y = 1 \mid X = 3)$.

Would you like to discuss this further or need additional clarification?

Here are 5 related questions:

1. How do you find $P(X = x)$ using joint probability values?
2. What does it mean for two variables to be independent in terms of their joint pmf?
3. Can you derive $P(Y = y)$ from a given joint pmf table?
4. What is the significance of marginal probabilities?
5. How would the solution change if $X$ and $Y$ were independent?

Tip: Conditional probability simplifies to multiplication if the events are independent. This is crucial when dealing with joint probability tables!