Question

Let \( X \) and \( Y \) be discrete random variables with the joint probability mass function \[ p(x, y) = \frac{1}{25} (x^2 + y^2), \quad \text{if} \, x = 1, 2; \, y = 0, 1, 2. \] Then \( P(Y = 1 | X = 1) \) equals

Hint:

For conditional probabilities with discrete random variables, use the joint probability mass function and the formula \( P(Y = 1 \, | \, X = 1) = \frac{P(X = 1, Y = 1)}{P(X = 1)} \).

Answer 1

Correct Answer: 0.25

Step 1: Understanding the problem.
We are given the joint probability mass function \( p(x, y) \) for discrete random variables \( X \) and \( Y \). The goal is to find \( P(Y = 1 \, | \, X = 1) \), the conditional probability of \( Y = 1 \) given that \( X = 1 \). The formula for conditional probability is: \[ P(Y = 1 \, | \, X = 1) = \frac{P(X = 1, Y = 1)}{P(X = 1)}. \]
Step 2: Calculating \( P(X = 1, Y = 1) \).
From the joint probability mass function, we calculate: \[ p(1, 1) = \frac{1}{25} (1^2 + 1^2) = \frac{1}{25} (1 + 1) = \frac{2}{25}. \]
Step 3: Calculating \( P(X = 1) \).
To calculate \( P(X = 1) \), sum over all possible values of \( Y \): \[ P(X = 1) = \sum_{y=0}^{2} p(1, y) = \frac{1}{25} \left( 1^2 + 0^2 \right) + \frac{1}{25} \left( 1^2 + 1^2 \right) + \frac{1}{25} \left( 1^2 + 2^2 \right) = \frac{1}{25} (1 + 2 + 5) = \frac{8}{25}. \]
Step 4: Calculating the conditional probability.
Now we can calculate \( P(Y = 1 \, | \, X = 1) \): \[ P(Y = 1 \, | \, X = 1) = \frac{\frac{2}{25}}{\frac{8}{25}} = \frac{2}{8} = 0.25. \]
Step 5: Conclusion.
Thus, \( P(Y = 1 \, | \, X = 1) = 0.25 \).