Question

A pair of fair dice is rolled. What is the probability that the second die lands on a higher value than the first?

• A. \( \frac{1}{36} \)
• B. \( \frac{5}{36} \)
• C. \( \frac{1}{6} \)
• D. \( \frac{5}{12} \)

Hint:

For ordered outcomes (e.g., dice rolling), list all possible cases systematically.

Answer 1

The Correct Option is D

Total outcomes when rolling two dice: \( 6 \times 6 = 36 \). 
Favorable outcomes where second die \( Y \) is greater than first die \( X \): \[ (1,2), (1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6), (4,5), (4,6), (5,6) \] Total favorable cases = 15. \[ P(Y>X) = \frac{15}{36} = \frac{5}{12} \] Conclusion: The probability is \( \frac{5}{12} \). 
 

Answer 2

Let the first die show a number \( x \) and the second die show a number \( y \), where \( x, y \in \{1, 2, 3, 4, 5, 6\} \). We are interested in the number of outcomes where \( y > x \).

The total number of outcomes when two dice are rolled is: \[ 6 \times 6 = 36 \]

Now, count the favorable outcomes where the second die is greater than the first:

• If first die is 1: second die can be 2,3,4,5,6 → 5 outcomes
• If first die is 2: second die can be 3,4,5,6 → 4 outcomes
• If first die is 3: second die can be 4,5,6 → 3 outcomes
• If first die is 4: second die can be 5,6 → 2 outcomes
• If first die is 5: second die can be 6 → 1 outcome
• If first die is 6: second die can’t be greater → 0 outcomes

Total favorable outcomes = \( 5 + 4 + 3 + 2 + 1 = 15 \) 

So, the probability is: \[ \frac{15}{36} = \frac{5}{12} \]

None of the given options are correct if the correct answer is \( \frac{5}{12} \).