Management Aptitude Test Sep 2020 Paper

Total number of integers $=20$
Number of ways in which 2 integers can be selected
$={}^{20}{\mathit{C}}_2=190$
For sum to be odd one integer must be odd and other even.
$\therefore$ Number of ways in which 1 odd and 1 even integer can be selected $=10\times 10=100$
$\therefore \mathit{P}($ sum is odd $)=\frac{100}{190}=\frac{10}{19}$