Question

If an unbiased dice is rolled thrice, then the probability of getting a greater number in the \( i \)-th roll than the number obtained in the \( (i-1) \)-th roll, \( i = 2, 3 \), is equal to:

• A. \( \frac{3}{54} \)
• B. \( \frac{2}{54} \)
• C. \( \frac{5}{54} \)
• D. \( \frac{1}{54} \)

Answer 1

The Correct Option is C

Favorable Cases:$\binom{6}{3}$

Total Outcomes: \(6^3\)

Probability of selecting numbers such that each is greater than the previous:

\[ P = \frac{\binom{6}{3}}{6^3} = \frac{20}{216} = \frac{5}{54} \]