Question:
Consider the sequence of numbers
\[
\{1,2,3,\dots,13\}.
\]
A person chooses three numbers at random from the sequence. The probability that the chosen three numbers form an A.P. is:
Consider the sequence of numbers
\[
\{1,2,3,\dots,13\}.
\]
A person chooses three numbers at random from the sequence. The probability that the chosen three numbers form an A.P. is:
Show Hint
For any sequence of consecutive integers from $1$ to $M$, if $M$ is odd, the number of triplets that form an A.P. can be found quickly using the shortcut formula: $\text{Number of A.P.s} = \frac{(M-1)^2}{4}$. Substituting $M = 13$ gives $\frac{(12)^2}{4} = \frac{144}{4} = 36$ favorable cases instantly!
Updated On: May 28, 2026
- $\frac{21}{157}$
- $\frac{18}{143}$
- $\frac{29}{180}$
- $\frac{24}{163}$
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
The total number of ways to select a sample of size $r$ from a larger population of size $N$ without replacement is given by the combination formula ${}^N C_r$. For three numbers $a, b, c$ to form an arithmetic progression, they must satisfy the condition $a + c = 2b$. This implies that the sum of the outer two numbers ($a+c$) must be an even number.
Step 1: Calculate the total number of possible sample combinations.
The total number of ways to choose 3 numbers at random from a set of 13 unique numbers is: $$\text{Total Outcomes } (N) = {}^{13}C_3 = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286$$
Step 2: Analyze the mathematical constraint for an A.P. triplet.
Let the three selected numbers be ordered such that $a < b < c$. For them to form an A.P., the spacing must be equal: $b - a = c - b \implies a + c = 2b$. Since $2b$ is always an even integer, the sum of the chosen numbers $a$ and $c$ must be even. A sum of two integers is even if and only if:
• Both numbers are odd ($\text{Odd} + \text{Odd} = \text{Even}$)
• Both numbers are even ($\text{Even} + \text{Even} = \text{Even}$) Once $a$ and $c$ are selected to satisfy this condition, the middle term $b$ is uniquely fixed as $b = \frac{a+c}{2}$.
Step 3: Count the number of favorable combinations.
Let's divide the set $\{1, 2, 3, \dots, 13\}$ into odd and even numbers:
• Odd numbers: $\{1, 3, 5, 7, 9, 11, 13\} \implies 7 \text{ numbers}$
• Even numbers: $\{2, 4, 6, 8, 10, 12\} \implies 6 \text{ numbers}$ The number of favorable ways to choose $a$ and $c$ is the number of ways to pick 2 numbers from the odd group, plus the number of ways to pick 2 numbers from the even group: $$\text{Favorable Outcomes } (n) = {}^7C_2 + {}^6C_2$$ $${}^7C_2 = \frac{7 \times 6}{2} = 21 \quad \text{and} \quad {}^6C_2 = \frac{6 \times 5}{2} = 15$$ $$n = 21 + 15 = 36$$
Step 4: Calculate the final probability.
$$\text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{n}{N} = \frac{36}{286} = \frac{18}{143}$$
The total number of ways to choose 3 numbers at random from a set of 13 unique numbers is: $$\text{Total Outcomes } (N) = {}^{13}C_3 = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286$$
Step 2: Analyze the mathematical constraint for an A.P. triplet.
Let the three selected numbers be ordered such that $a < b < c$. For them to form an A.P., the spacing must be equal: $b - a = c - b \implies a + c = 2b$. Since $2b$ is always an even integer, the sum of the chosen numbers $a$ and $c$ must be even. A sum of two integers is even if and only if:
• Both numbers are odd ($\text{Odd} + \text{Odd} = \text{Even}$)
• Both numbers are even ($\text{Even} + \text{Even} = \text{Even}$) Once $a$ and $c$ are selected to satisfy this condition, the middle term $b$ is uniquely fixed as $b = \frac{a+c}{2}$.
Step 3: Count the number of favorable combinations.
Let's divide the set $\{1, 2, 3, \dots, 13\}$ into odd and even numbers:
• Odd numbers: $\{1, 3, 5, 7, 9, 11, 13\} \implies 7 \text{ numbers}$
• Even numbers: $\{2, 4, 6, 8, 10, 12\} \implies 6 \text{ numbers}$ The number of favorable ways to choose $a$ and $c$ is the number of ways to pick 2 numbers from the odd group, plus the number of ways to pick 2 numbers from the even group: $$\text{Favorable Outcomes } (n) = {}^7C_2 + {}^6C_2$$ $${}^7C_2 = \frac{7 \times 6}{2} = 21 \quad \text{and} \quad {}^6C_2 = \frac{6 \times 5}{2} = 15$$ $$n = 21 + 15 = 36$$
Step 4: Calculate the final probability.
$$\text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{n}{N} = \frac{36}{286} = \frac{18}{143}$$
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