Question

Consider the sample space \( \Omega = \{(x, y): x, y \in \{1, 2, 3, 4\} \} \) where each outcome is equally likely. Let \( A = \{ x \geq 2 \} \) and \( B = \{ y > x \} \) be two events. Then which of the following is NOT true?

• A. \( A \cap B = \frac{1}{4} \)
• B. \( A \) and \( B \) are not independent
• C. \( P(B) = \frac{3}{8} \)
• D. \( P(A) = \frac{3}{4} \)

Hint:

To check independence of events, verify if \( P(A \cap B) = P(A) \times P(B) \). If not, the events are not independent.

Answer 1

The Correct Option is A

Step 1: The sample space \( \Omega \) consists of all ordered pairs \( (x, y) \), where \( x, y \in \{1, 2, 3, 4\} \). Thus, the total number of outcomes is: \[ |\Omega| = 4 \times 4 = 16 \] Step 2: Event \( A \) is defined as \( A = \{x \geq 2\} \), which includes the pairs where \( x \) is 2, 3, or 4. Therefore, the number of outcomes in \( A \) is: \[ |A| = 3 \times 4 = 12 \] Thus, \( P(A) = \frac{12}{16} = \frac{3}{4} \). Step 3: Event \( B \) is defined as \( B = \{y > x\} \). We now count the outcomes where \( y > x \) for each possible value of \( x \): - For \( x = 1 \), \( y \in \{2, 3, 4\} \) (3 outcomes). - For \( x = 2 \), \( y \in \{3, 4\} \) (2 outcomes). - For \( x = 3 \), \( y \in \{4\} \) (1 outcome). - For \( x = 4 \), there are no values of \( y \) greater than 4. Thus, the total number of outcomes in \( B \) is: \[ |B| = 3 + 2 + 1 = 6 \] Therefore, \( P(B) = \frac{6}{16} = \frac{3}{8} \). Step 4: The intersection of events \( A \cap B \) corresponds to outcomes where both \( x \geq 2 \) and \( y > x \). These outcomes are: - For \( x = 2 \), \( y \in \{3, 4\} \) (2 outcomes). - For \( x = 3 \), \( y = 4 \) (1 outcome). - For \( x = 4 \), there are no outcomes where \( y > x \). Thus, the total number of outcomes in \( A \cap B \) is: \[ |A \cap B| = 2 + 1 = 3 \] Therefore, \( P(A \cap B) = \frac{3}{16} \). Step 5: Since \( P(A \cap B) = \frac{3}{16} \) and not \( \frac{1}{4} \), option (a) is incorrect.