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

To solve this problem, we need to determine the probability of obtaining a greater number with each successive dice roll when an unbiased dice is rolled thrice. Let's break down the problem step-by-step.

The number of total possible outcomes when a dice is rolled thrice is:

\(6^3 = 216\)

Now, we need to consider favorable cases where the number on the second roll is greater than the first, and the number on the third roll is greater than the second. Let's consider these rolls:

First roll (\( x_1 \)): Any number from 1 to 6 can appear. There are 6 choices for this.

Second roll (\( x_2 \)): It must be greater than \( x_1 \). So, if \( x_1 \) is 1, then \( x_2 \) can be 2, 3, 4, 5, or 6. Similarly, if \( x_1 \) is 2, \( x_2 \) can be 3, 4, 5, or 6, and so on.

Third roll (\( x_3 \)): It must be greater than \( x_2 \) following the same pattern. For instance, if \( x_2 \) is 2, then \( x_3 \) can be 3, 4, 5, or 6.

For example, consider when:

• If \( x_1 = 1 \), then \( x_2 \) could be 2, 3, 4, 5, or 6 (5 possibilities), and for each \( x_2 \), \( x_3 \) has respective possibilities.
• If \( x_1 = 2 \), then \( x_2 \) could be 3, 4, 5, or 6 (4 possibilities), and for each \( x_2 \), \( x_3 \) has respective possibilities.

We calculate the sum of all such cases:

\(\begin{aligned} &1: 5 \, \text{(five choices for } x_2) \\ &\quad\quad \to 4 + 3 + 2 + 1 \, \text{ (these are the choices for } x_3) \\ &2: 4 \, \text{(four choices for } x_2) \\ &\quad\quad \to 3 + 2 + 1 \, \text{ (these are the choices for } x_3) \\ &3: 3 \, \text{(three choices for } x_2) \\ &\quad\quad \to 2 + 1 \\ &4: 2 \, \text{(two choices for } x_2) \\ &\quad\quad \to 1 \\ \end{aligned}\)

The calculated number of favorable outcomes is the sum of these numbers then:

\(10 + 6 + 3 + 1 = 20\)

Therefore, the probability that the number on the \(i\)-th roll is greater than the number on the \((i-1)\)-th roll for \(i = 2, 3\) is:

\(\frac{20}{216} = \frac{5}{54}\)

Thus, the correct option is \(\frac{5}{54}\).

Answer 2

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

Total Outcomes: \(6^3\)

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

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