Question:
3 numbers are selected randomly from numbers \(1,2,3,\ldots,31\). The probability that they are in A.P. is
3 numbers are selected randomly from numbers \(1,2,3,\ldots,31\). The probability that they are in A.P. is
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For three numbers to form an A.P., the middle term must be the average of the first and third terms.
Updated On: Apr 9, 2026
- \(\frac{15}{31}\)
- \(\frac{7}{31}\)
- \(\frac{8}{17}\)
- \(\frac{45}{899}\)
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The Correct Option is D
Solution and Explanation
Concept:
Three numbers \(a,b,c\) are in arithmetic progression if
\[
b-a = c-b
\]
which implies
\[
a+c = 2b
\]
Thus \(b\) must be the average of \(a\) and \(c\).
Step 1: Total number of ways Selecting any \(3\) numbers from \(31\): \[ ^{31}C_3=\frac{31\times30\times29}{3\times2\times1} \] \[ =4495 \]
Step 2: Condition for A.P. For \(b\) to be the average of \(a\) and \(c\), \(a\) and \(c\) must both be either odd or even. From numbers \(1\) to \(31\): Odd numbers \(=16\) Even numbers \(=15\)
Step 3: Choose pairs Pairs of odd numbers: \[ ^{16}C_2=120 \] Pairs of even numbers: \[ ^{15}C_2=105 \] Total favourable selections: \[ 120+105=225 \]
Step 4: Probability \[ P=\frac{225}{4495} \] \[ P=\frac{45}{899} \]
Step 1: Total number of ways Selecting any \(3\) numbers from \(31\): \[ ^{31}C_3=\frac{31\times30\times29}{3\times2\times1} \] \[ =4495 \]
Step 2: Condition for A.P. For \(b\) to be the average of \(a\) and \(c\), \(a\) and \(c\) must both be either odd or even. From numbers \(1\) to \(31\): Odd numbers \(=16\) Even numbers \(=15\)
Step 3: Choose pairs Pairs of odd numbers: \[ ^{16}C_2=120 \] Pairs of even numbers: \[ ^{15}C_2=105 \] Total favourable selections: \[ 120+105=225 \]
Step 4: Probability \[ P=\frac{225}{4495} \] \[ P=\frac{45}{899} \]
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