1. Key Concepts and Formulas

This problem requires us to calculate the probability of an event using combinations. Let's first define the fundamental concepts:

• Probability: The probability of an event $E$ is the ratio of the number of favorable outcomes to the total number of possible outcomes. $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
• Combinations: When selecting a subset of items from a larger set where the order of selection does not matter, we use combinations. The number of ways to choose $k$ distinct items from a set of $n$ distinct items is given by the combination formula: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$
• Arithmetic Progression (A.P.): Three numbers $a, b, c$ are said to be in Arithmetic Progression if the difference between consecutive terms is constant. This means $b-a = c-b$, which simplifies to $2b = a+c$. For the numbers to be in A.P. with a positive common difference, we must have $a < b < c$.

2. Step-by-Step Solution

Step 1: Identify the set of numbers and their properties. We are given 11 consecutive natural numbers. For calculation purposes, we can assume this set is $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$. The specific starting number for consecutive natural numbers does not affect the distribution of odd and even numbers, nor the counts of possible APs. From this set, we identify the odd and even numbers:

• Odd numbers (O): $\{1, 3, 5, 7, 9, 11\}$. There are 6 odd numbers.
• Even numbers (E): $\{2, 4, 6, 8, 10\}$. There are 5 even numbers.

Step 2: Calculate the total number of possible outcomes. We need to select three numbers at random (without repetition) from the set of 11 consecutive natural numbers. Since the order of selection does not matter, we use combinations. The total number of ways to choose 3 numbers from 11 is: $\text{Total outcomes} = \binom{11}{3}$ $\binom{11}{3} = \frac{11!}{3!(11-3)!} = \frac{11!}{3!8!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1}$ $\binom{11}{3} = 11 \times 5 \times 3 = 165$ So, there are 165 possible ways to select three numbers.

Step 3: Calculate the number of favorable outcomes. We are looking for selections of three numbers $(a, b, c)$ such that they are in A.P. with a positive common difference. This means $a < b < c$ and $2b = a+c$. The condition $2b = a+c$ implies that $a+c$ must be an even number. For the sum of two integers to be even, both integers must have the same parity (i.e., both must be odd, or both must be even).

We consider two cases for the parity of $a$ and $c$:

• Case 1: Both $a$ and $c$ are odd numbers. If $a$ and $c$ are both odd, their sum $a+c$ will be even, and thus $b = (a+c)/2$ will be an integer. Since $a < c$, we have $a < (a+c)/2 < c$, so $b$ will naturally fall between $a$ and $c$. The common difference $d = b-a = (c-a)/2$ will be a positive integer because $a$ and $c$ are distinct odd numbers. All such $b$ values will be within the range $[(1+3)/2, (9+11)/2] = [2, 10]$, which are all within our set $S$. The number of odd numbers available is 6. The number of ways to choose 2 distinct odd numbers from 6 is: $\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15$ These 15 pairs correspond to 15 unique A.P.s (e.g., choosing $\{1, 3\}$ gives $(1,2,3)$; choosing $\{1, 5\}$ gives $(1,3,5)$, etc.).

• Case 2: Both $a$ and $c$ are even numbers. Similarly, if $a$ and $c$ are both even, their sum $a+c$ will be even, and $b = (a+c)/2$ will be an integer. As before, $b$ will be between $a$ and $c$, and the common difference will be a positive integer. All such $b$ values will be within the range $[(2+4)/2, (8+10)/2] = [3, 9]$, which are all within our set $S$. The number of even numbers available is 5. The number of ways to choose 2 distinct even numbers from 5 is: $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$ These 10 pairs correspond to 10 unique A.P.s (e.g., choosing $\{2, 4\}$ gives $(2,3,4)$; choosing $\{2, 6\}$ gives $(2,4,6)$, etc.).

The total number of favorable outcomes (A.P.s with positive common difference) is the sum of the outcomes from Case 1 and Case 2: $\text{Favorable outcomes} = 15 + 10 = 25$

Step 4: Calculate the probability. Now we can calculate the probability by dividing the number of favorable outcomes by the total number of outcomes: $P(\text{A.P.}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{25}{165}$ To simplify the fraction, we can divide both the numerator and the denominator by their greatest common divisor, which is 5: $P(\text{A.P.}) = \frac{25 \div 5}{165 \div 5} = \frac{5}{33}$

3. Common Mistakes & Tips

• Parity is key for AP: The property $2b = a+c$ is crucial. It immediately tells us that $a$ and $c$ must have the same parity. This is a very efficient way to count favorable outcomes. Trying to list APs by common difference can be more tedious and error-prone for larger sets.
• "Positive common difference": This implies $a < b < c$. Our method of choosing two distinct numbers $a$ and $c$ and then determining $b$ automatically ensures this order and a positive common difference.
• Don't forget to simplify: Always simplify the final probability fraction to its lowest terms.

4. Summary

To find the probability, we first determined the total number of ways to select 3 numbers from 11 using combinations, which is $\binom{11}{3} = 165$. Next, we identified the conditions for three numbers to form an A.P. with a positive common difference ($a < b < c$ and $2b = a+c$). This condition implies that the first and third terms ($a$ and $c$) must have the same parity. We counted the favorable outcomes by considering cases where $a$ and $c$ are both odd (15 ways) and where $a$ and $c$ are both even (10 ways), leading to a total of 25 favorable outcomes. Finally, the probability was calculated as the ratio of favorable outcomes to total outcomes, $\frac{25}{165} = \frac{5}{33}$.

The final answer is $\boxed{\text{(A)}}$.