1. Key Concepts and Formulas

• Classical Probability Definition: For an event $E$ in a finite sample space of equally likely outcomes, the probability $P(E)$ is given by: $P(E) = \frac{\text{Number of Favorable Outcomes for E}}{\text{Total Number of Possible Outcomes}}$
• Fundamental Principle of Counting: If an event can occur in $m$ ways and another independent event can occur in $n$ ways, then the two events can occur in $m \times n$ ways. This principle is used to determine the total number of possible outcomes when multiple independent trials (like rolling a die multiple times) are performed.
• Combinations: The number of ways to choose $r$ distinct items from a set of $n$ distinct items, where the order of selection does not matter, is given by the combination formula: $^n C_r = \frac{n!}{r!(n-r)!}$ This is particularly useful when the problem conditions implicitly fix the order of the selected items.

2. Step-by-Step Solution

Step 1: Define the Event and Determine the Total Number of Possible Outcomes

• What we are doing: We first need to precisely understand the event whose probability we are calculating and then determine the size of the entire sample space (all possible outcomes).

• Why: This calculation will form the denominator of our probability fraction.

• Let $X_1$, $X_2$, and $X_3$ represent the numbers obtained on the first, second, and third rolls of an unbiased six-sided die, respectively. The set of possible outcomes for each individual roll is $\{1, 2, 3, 4, 5, 6\}$.

• The problem states that we need to find the probability of "getting a greater number in the $i^{\text{th}}$ roll than the number obtained in the $(i-1)^{\text{th}}$ roll, for $i=2,3$." Let's break down these conditions:

  • For $i=2$: This means the number in the second roll ($X_2$) must be greater than the number in the first roll ($X_1$). Mathematically, $X_2 > X_1$.
  • For $i=3$: This means the number in the third roll ($X_3$) must be greater than the number in the second roll ($X_2$). Mathematically, $X_3 > X_2$.

• Since both conditions must hold simultaneously, we are looking for the probability of the event where $X_1 < X_2 < X_3$.

• Calculation of Total Outcomes:

  • For the first roll ($X_1$), there are 6 possible outcomes.
  • For the second roll ($X_2$), there are 6 possible outcomes.
  • For the third roll ($X_3$), there are 6 possible outcomes.
  • Since each roll is an independent event, the total number of distinct ordered sequences of outcomes $(X_1, X_2, X_3)$ is found by multiplying the number of outcomes for each individual roll using the Fundamental Principle of Counting: $\text{Total Number of Outcomes} = 6 \times 6 \times 6 = 6^3 = 216$

Step 2: Determine the Number of Favorable Outcomes

• What we are doing: We need to count the number of sequences $(X_1, X_2, X_3)$ that satisfy the specific condition $X_1 < X_2 < X_3$.
• Why: This count will form the numerator of our probability fraction.
• The condition $X_1 < X_2 < X_3$ implies that the three numbers obtained from the rolls must be distinct and arranged in strictly increasing order.
• Consider the set of possible outcomes for a single roll: $\{1, 2, 3, 4, 5, 6\}$.
• If we choose any three distinct numbers from this set, there is only one unique way to arrange them such that they satisfy the strictly increasing condition $X_1 < X_2 < X_3$. For example, if we choose the numbers $\{2, 5, 3\}$, the only arrangement that satisfies the condition is $(2, 3, 5)$. The order is fixed once the numbers are chosen.
• Therefore, the problem simplifies to simply selecting 3 distinct numbers from the 6 available faces of the die. This is a classic combinatorial problem solved using the combination formula $^n C_r$.
• Calculation of Favorable Outcomes:
  • Here, $n=6$ (the 6 faces of the die) and $r=3$ (the 3 numbers that form our sequence).
  • Using the combination formula $^n C_r$: $\text{Number of Favorable Outcomes} = ^6 C_3 = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!}$ $= \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(3 \times 2 \times 1)} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 5 \times 4 = 20$
• To illustrate and confirm, here are some of the 20 favorable outcomes: $(1,2,3), (1,2,4), \dots, (1,5,6), (2,3,4), \dots, (4,5,6)$.

Step 3: Calculate the Probability and Simplify

• What we are doing: We apply the classical probability definition using the total number of outcomes and the number of favorable outcomes calculated in the previous steps.
• Why: To obtain the final answer.
• Using the values determined in Step 1 and Step 2:
  • Number of Favorable Outcomes = 20
  • Total Number of Outcomes = 216 $P(X_1 < X_2 < X_3) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Outcomes}} = \frac{20}{216}$
• Simplification: We simplify the fraction to its lowest terms by dividing both the numerator and denominator by their greatest common divisor (GCD).
  • Both numbers are divisible by 4: $20 \div 4 = 5$ and $216 \div 4 = 54$.
  • The simplified probability is $\frac{5}{54}$.

3. Common Mistakes & Tips

• Misinterpreting the Condition: Carefully read and understand the problem statement. The phrase "getting a greater number in the $i^{\text{th}}$ roll than the number obtained in the $(i-1)^{\text{th}}$ roll, $i=2,3$" explicitly means $X_2 > X_1$ AND $X_3 > X_2$, which simplifies to $X_1 < X_2 < X_3$. A common mistake is to interpret "or" instead of "and", or to use non-strict inequalities ($\le$) instead of strict inequalities ($<$).
• Combinations vs. Permutations: This is a crucial distinction. Since the condition $X_1 < X_2 < X_3$ fixes the order of the chosen numbers, we use combinations ($^n C_r$) to select the numbers. If the order were not fixed (e.g., if we just needed three distinct numbers in any order), then permutations ($^n P_r$) would be relevant for counting ordered sequences.
• Simplifying Fractions: Always reduce the final probability fraction to its lowest terms. This is a standard practice in competitive examinations and ensures the answer is in the expected format.

4. Summary

To solve this problem, we first established that the total number of possible outcomes when rolling an unbiased die thrice is $6^3 = 216$. The given condition, $X_i > X_{i-1}$ for $i=2,3$, translates to requiring a strictly increasing sequence of numbers, i.e., $X_1 < X_2 < X_3$. This means we need to choose 3 distinct numbers from the 6 possible outcomes, and their order is automatically determined by the strictly increasing condition. Using combinations, the number of favorable outcomes was calculated as $^6 C_3 = 20$. Finally, the probability was found by dividing the number of favorable outcomes by the total number of outcomes, yielding $\frac{20}{216}$, which simplifies to $\frac{5}{54}$.

5. Final Answer

The final answer is $\boxed{\frac{5}{54}}$, which corresponds to option (A).