Question

Consider four dice \( D_1, D_2, D_3, \) and \( D_4 \), each having six faces marked as follows:

Die	Marks on faces
\(D_1\)	4, 4, 4, 4, 0, 0
\(D_2\)	3, 3, 3, 3, 3, 3
\(D_3\)	6, 6, 2, 2, 2, 2
\(D_4\)	5, 5, 5, 1, 1, 1

In each roll of a die, each of its six faces is equally likely to occur. Suppose that each of these four dice is rolled once, and the marks on their upper faces are noted. Let the four rolls be independent. If \( X_i \) denotes the mark on the upper face of the die \( D_i \), \( i = 1, 2, 3, 4 \), then which of the following statements is/are correct?

• A. \( P(X_1 > X_2) = \frac{2}{3} \)
• B. \( P(X_3 > X_4) = \frac{2}{3} \)
• C. \( P(X_2 > X_3) = \frac{1}{3} \)
• D. The events \(\{X_1 > X_2\}\) and \(\{X_2 > X_3\}\) are independent

Answer 1

The Correct Option is A, B, D

1. Dice \( D_1 \) and \( D_2 \): - Dice \( D_1 \) has the outcomes \( \{4, 4, 4, 4, 0, 0\} \), each with probability \( \frac{1}{6} \). - Dice \( D_2 \) has the outcomes \( \{3, 3, 3, 3, 3, 3\} \), each with probability \( \frac{1}{6} \). - To compute \( P(X_1 > X_2) \), consider: \[ P(X_1 = 4 \text{ and } X_2 = 3) = \frac{4}{6} \times \frac{6}{6} = \frac{4}{6}. \] - Thus: \[ P(X_1 > X_2) = \frac{4}{6} = \frac{2}{3}. \] 2. Dice \( D_3 \) and \( D_4 \): - Similar computations show \( P(X_3 > X_4) \neq \frac{2}{3} \) due to the different combinations of outcomes. 3. Correctness of Option (A): - From the above analysis, \( P(X_1 > X_2) = \frac{2}{3} \). 4. Independence of Events: - The events \( \{X_1 > X_2\} \) and \( \{X_2 > X_3\} \) are not independent, as they depend on overlapping variables