Calculating Arithmetic Progression Probability

The problem asks for the probability that three numbers chosen randomly from the set $S = \{1, 2, 3, ..., 13\}$ form an arithmetic progression (A.P.).

Total Possible Outcomes

The total number of ways to choose 3 distinct numbers from the set of 13 numbers is given by the combination formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).

Total combinations = \(\binom{13}{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 13 \times 2 \times 11 = 286\). So, there are 286 possible ways to choose 3 numbers.

Favorable Outcomes: Arithmetic Progressions

A set of three distinct numbers $\{x, y, z\}$ forms an arithmetic progression if, when sorted in increasing order ($x < y < z$), the difference between consecutive terms is constant, i.e., $y - x = z - y$. This condition is equivalent to $x + z = 2y$.

For $x + z = 2y$ to hold with integers $x, y, z$, the sum $x + z$ must be an even number. This implies that $x$ and $z$ must have the same parity (both must be odd or both must be even).

We need to count the number of pairs $(x, z)$ from the set $S$ such that $x < z$ and both $x, z$ have the same parity.

• The set $S = \{1, 2, ..., 13\}$ contains 7 odd numbers: $\{1, 3, 5, 7, 9, 11, 13\}$.
• The set $S$ contains 6 even numbers: $\{2, 4, 6, 8, 10, 12\}$.

Case 1: Both $x$ and $z$ are odd. The number of ways to choose 2 distinct odd numbers from the 7 available odd numbers is \(\binom{7}{2}\). \(\binom{7}{2} = \frac{7 \times 6}{2} = 21\). Each such pair $(x, z)$ corresponds to a unique A.P. triplet $(x, (x+z)/2, z)$ where $x, z \in S$, $x$ is odd, $z$ is odd, $x < z$, and the middle term $y=(x+z)/2$ is an integer within $S$. There are 21 such triplets.

Case 2: Both $x$ and $z$ are even. The number of ways to choose 2 distinct even numbers from the 6 available even numbers is \(\binom{6}{2}\). \(\binom{6}{2} = \frac{6 \times 5}{2} = 15\). Each such pair $(x, z)$ corresponds to a unique A.P. triplet $(x, (x+z)/2, z)$ where $x, z \in S$, $x$ is even, $z$ is even, $x < z$, and the middle term $y=(x+z)/2$ is an integer within $S$. There are 15 such triplets.

Total number of A.P. triplets = (Number of triplets from odd pairs) + (Number of triplets from even pairs) Total A.P.s = $21 + 15 = 36$. There are 36 distinct sets of three numbers from $\{1, ..., 13\}$ that form an arithmetic progression.

Calculating the Probability

The probability is the ratio of the number of favorable outcomes (A.P.s) to the total number of possible outcomes.

Probability = \(\frac{\text{Number of A.P. triplets}}{\text{Total number of triplets}} = \frac{36}{286}\).

Simplifying the fraction:

\(\frac{36}{286} = \frac{36 \div 2}{286 \div 2} = \frac{18}{143}\).

The probability that the chosen three numbers form an A.P. is $\frac{18}{143}$.