Question

Out of $31$ numbers $1,2,3,...,31$, three numbers are chosen at random, without replacement. Let $p$ denote the probability that the selected numbers are in A.P. find $\frac{899}{p}$.

Answer 1

Any three numbers out of $31$ van be chosen in ${}^{31}{C}_3$ ways.
For the favourable choice if the chosen numbers are $a,b$ and $c$, then $\frac{a+c}{2}=b$.
Obviously either both $a$ and $c$ are even or both are odd and then $b$ is fixed.
Hence for favourable choice we have to chose two numbers from $1$ to $31$ in
${}^{16}{C}_2{+}^{15}{C}_2$ ways.
Hence required probability is $\frac{{}^{16}{C}_2{+}^{15}{C}_2}{{}^{31}{C}_3}=\frac{8\times 15+15\times 7}{5\times 31\times 29}=\frac{45}{899}$