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 $\frac{5}{12}$.