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

Solution

Maths sir · 4 steps

Pick any three different faces: there is exactly one way to roll them in increasing order.

1. Total outcomes. Each roll has 6 outcomes, so three rolls give 6³ = 216 equally likely ordered triples (x₁, x₂, x₃).
2. What is needed. The condition for i = 2, 3 says x₂ > x₁ and x₃ > x₂, so x₁ < x₂ < x₃. All three numbers must be different and in strictly increasing order.
3. Count favourable. Choose any 3 different numbers from {1, 2, 3, 4, 5, 6}. Only one arrangement of them is increasing, so each choice gives exactly one favourable triple. That is C(6, 3) = 20 triples.

   P = C(6, 3)/6³ = 20/216 = 5/54

4. So. The probability is 5/54, which is option (a).

The idea to keep. A strictly increasing sequence is fixed once you choose its values, so count it with C(n, r), not by listing cases. If repeats were allowed (non-decreasing), you would need a different count.

Answer from the NTA final answer key. Solution by PrepWiser.