Question

Out of tickets marked with numbers from $1$ to $21$, three are drawn at random, the probability that the three numbers on them are in A.P. is $\frac{2k}{133}$. Find the value of $k$.

Answer 1

Suppose $3$ no's are $a,b,c$.
Therefore $2b=a+c$ or $a+c=$ even
$\Rightarrow a$ and $c$ are both even or both odd
Therefore favorable number of ways ${}^{10}{C}_2{+}^{11}{C}_2=100$
Since there are $10$ even and $11$ odd numbers
Sample space ${=}^{21}{C}_2=1330$
Hence required probability $=\frac{100}{1330}=\frac{10}{133}$