AIEEE 2010 (Offline)Question 1
Four numbers are chosen at random (without replacement) from the set $\{1,2,3,\ldots,20\}$.
Statement - 1: The probability that the chosen numbers when arranged in some order will form an AP is $\frac{1}{85}$.
Statement - 2: If the four chosen numbers form an AP, then the set of all possible values of common difference is $(\pm1,\pm2,\pm3,\pm4,\pm5)$.
Probability
Solution
Common difference $d$ gives $20-3d$ progressions for $d=1,\dots,6$: $17+14+11+8+5+2=57$, and $\frac{57}{\binom{20}{4}}=\frac1{85}$. So Statement 1 is true, but $d=\pm6$ is also possible (1,7,13,19), so Statement 2 is false.