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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!

Learn how to calculate conditional probability P(Y = 1 | X = 3) using joint probability tables and marginal probabilities. This problem explores key concepts in probability theory, including conditional probability and joint probability. Suitable for college-level students and those interested in advanced probability calculations.

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