Question

A die is rolled three times. The probability of getting a larger number than the previous number is

• A. \(\frac{5}{216}\)
• B. \(\frac{5}{54}\)
• C. \(\frac{1}{6}\)
• D. \(\frac{5}{36}\)

Hint:

Strictly increasing sequences = combinations \({}^nC_r\).

Answer 1

The Correct Option is B

To solve this problem, we need to find the probability that we will roll three dice in sequence, and each roll will result in a higher number than the previous roll. Let's break this down step by step.

1. The total number of outcomes when rolling a die three times is given by:
   \(6 \times 6 \times 6 = 216\).
2. Next, we need to find the number of favorable outcomes where each subsequent number rolled is greater than the previous number.
3. Consider the sequence of rolls. Since each number must be greater than the previous, the numbers must all be different and ordered in increasing fashion. This is akin to selecting 3 distinct numbers from 6 and arranging them in increasing order - there is only one way to do this for any given set of numbers.
4. Selecting 3 distinct numbers from 6 can be done in \(\binom{6}{3}\) ways, which is calculated as follows:
   \(\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20\).
5. Since each selection of 3 numbers results in exactly one valid sequence, the number of favorable outcomes is 20.
6. Therefore, the probability of rolling the dice such that each number is greater than the previous one is:
   \(\frac{20}{216} = \frac{5}{54}\).

The correct answer is \(\frac{5}{54}\), which matches with the provided correct option.