Question

Out of 21 tickets marked with number 1 to 21, 3 tickets are drawn at random. Find the probability that the 3 numbers on them are in A.P.
if it is expressed in simplest form as $a/b$, $a+b=?$

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 $=\frac{100}{1330}=\frac{10}{133}$