Question

Two dice are thrown. Defined are the following two events A and B: \[ A = \{(x, y) : x + y = 9\}, \quad B = \{(x, y) : x \neq 3\}, \] where \( (x, y) \) denote a point in the sample space. Check if events \( A \) and \( B \) are independent or mutually exclusive.

Hint:

Events \( A \) and \( B \) are independent if \( P(A \cap B) = P(A)P(B) \) and mutually exclusive if \( P(A \cap B) = 0 \).

Answer 1

Step 1: Compute \( P(A) \). The total sample space for rolling two dice is \( 6 \times 6 = 36 \). Event \( A \) consists of pairs where \( x + y = 9 \): \[ (3,6), (4,5), (5,4), (6,3). \] So, \[ P(A) = \frac{4}{36} = \frac{1}{9}. \] Step 2: Compute \( P(B) \). Event \( B \) consists of all outcomes where \( x \neq 3 \), meaning that \( x \) can take values \( \{1,2,4,5,6\} \) (5 choices for \( x \), each paired with 6 possible \( y \) values): \[ \text{Total favorable outcomes} = 5 \times 6 = 30. \] Thus, \[ P(B) = \frac{30}{36} = \frac{5}{6}. \] Step 3: Compute \( P(A \cap B) \). Find outcomes in both \( A \) and \( B \): \[ A = \{(3,6), (4,5), (5,4), (6,3)\}. \] Since \( B \) excludes outcomes where \( x = 3 \), the valid outcomes are: \[ (4,5), (5,4), (6,3). \] Thus, \[ P(A \cap B) = \frac{3}{36} = \frac{1}{12}. \] Step 4: Check Independence. Events \( A \) and \( B \) are independent if: \[ P(A \cap B) = P(A) \cdot P(B). \] Computing: \[ \frac{1}{12} \neq \frac{1}{9} \times \frac{5}{6} = \frac{5}{54}. \] Since \( P(A \cap B) \neq P(A)P(B) \), events \( A \) and \( B \) are not independent. Step 5: Check Mutual Exclusivity. Events are mutually exclusive if \( P(A \cap B) = 0 \). Since \( P(A \cap B) = \frac{1}{12} \neq 0 \), \( A \) and \( B \) are not mutually exclusive. Final Conclusion: \( A \) and \( B \) are neither independent nor mutually exclusive.