Step 1: Understanding the Concept:
For three numbers \( a, b, c \) (in increasing order) to form an A.P., the condition is \( a + c = 2b \). This implies that the sum of the first and last number must be an even number. This happens if \( a \) and \( c \) are either both even or both odd. The middle term \( b \) is then automatically determined as \( (a+c)/2 \).
Step 2: Key Formula or Approach:
1. Total outcomes: Choosing 3 numbers from 13: \( \binom{13}{3} \).
2. Favorable outcomes: Choosing 2 numbers with the same parity from the set.
Step 3: Detailed Explanation:
Total sample space:
\[ n(S) = \binom{13}{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286 \]
For three numbers to be in A.P., we only need to pick the endpoints \( a \) and \( c \) such that their sum is even.
Number of Odd numbers in \( \{1, 2, \dots, 13\} \) is 7: \( \{1, 3, 5, 7, 9, 11, 13\} \).
Number of Even numbers in \( \{1, 2, \dots, 13\} \) is 6: \( \{2, 4, 6, 8, 10, 12\} \).
Case 1: Both endpoints are Odd.
Ways \( = \binom{7}{2} = 21 \).
Case 2: Both endpoints are Even.
Ways \( = \binom{6}{2} = 15 \).
Total favorable outcomes \( = 21 + 15 = 36 \).
Each pair \( (a, c) \) uniquely determines \( b \). Since \( a \) and \( c \) are chosen from the set and their sum is even, \( b = (a+c)/2 \) will always be an integer and will always lie between \( a \) and \( c \), hence \( b \) is also in the set.
Probability \( P = \frac{36}{286} \).
Reducing the fraction: \( \frac{18}{143} \).
Step 4: Final Answer:
The probability is calculated by dividing the 36 possible A.P. sets by the total 286 combinations, resulting in \( 18/143 \).