If two numbers are selected randomly from 20 consecutive natural numbers, find the probability that the sum of the two numbers is (i) an odd number (ii) an even number.

i)   Two numbers are selected from 1 – 20 natural numbers 

 $n\left(S\right){=}^{20}{C}_2$

ii) Let $E$ be the event of selecting that sum of two numbers is odd, 

even no’s $=10$ , odd no’s $=10$

$n\left(E\right){=}^{10}{C}_1{\cdot }^{10}{C}_1,$  $n\left(S\right){=}^{20}{C}_2$

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{10\times 10}{{{20}_{C_2}}_{ }}=\frac{10\times 10}{\frac{20\times 19}{1\times 2}}$

$=\frac{10\times 10\times 2}{20\times 19}=\frac{10}{19}$

Probability of selecting two numbers sum is odd$=10/19$

iii) Probability of selecting two numbers whose sum is even $=1-\frac{10}{19}=\frac{9}{19}$