Question

Let \( X \) and \( Y \) have the joint probability mass function \[ P(X = m, Y = n) = \begin{cases} \frac{m + n}{21}, & m = 1, 2, 3; n = 1, 2, \\ 0, & \text{otherwise}. \end{cases}\] Then \( P(X = 2 | Y = 2) \) equals 
 

• A. \( \dfrac{1}{3} \)
• B. \( \dfrac{2}{3} \)
• C. \( \dfrac{1}{2} \)
• D. \( \dfrac{1}{4} \)

Hint:

When calculating conditional probabilities, use the formula \( P(X | Y) = \frac{P(X, Y)}{P(Y)} \).

Answer 1

The Correct Option is A

Step 1: Find the conditional probability formula. 
The conditional probability \( P(X = 2 | Y = 2) \) is given by: \[ P(X = 2 | Y = 2) = \frac{P(X = 2, Y = 2)}{P(Y = 2)}. \]

Step 2: Find \( P(X = 2, Y = 2) \). 
From the given joint mass function: \[ P(X = 2, Y = 2) = \frac{2 + 2}{21} = \frac{4}{21}. \]

Step 3: Find \( P(Y = 2) \). 
To find \( P(Y = 2) \), sum the joint probabilities for all \( X \) values when \( Y = 2 \): \[ P(Y = 2) = \frac{1 + 2}{21} + \frac{2 + 2}{21} + \frac{3 + 2}{21} = \frac{3}{21} + \frac{4}{21} + \frac{5}{21} = \frac{12}{21}. \]

Step 4: Calculate the conditional probability. 
Now, calculate \( P(X = 2 | Y = 2) \): \[ P(X = 2 | Y = 2) = \frac{\frac{4}{21}}{\frac{12}{21}} = \frac{4}{12} = \frac{1}{3}. \]

Step 5: Conclusion. 
The correct answer is (A) \( \frac{1}{3} \).