If a proper divisior of the integer 2520 is selected at random, then the probability that it is an odd number is 

$\begin{array}{rcl}2520& =& 2\times 126\times 5\times 2\\ & =& 2^2\times 2\times +63\times 5\\ & =& 2^3\times 7\times 3^2\times 5\\ \mathrm{n}\left(\mathrm{S}\right)& =& (3+1) (1+1) (2+1) (1+1)=48\end{array}$

n(S) = number  of proper divisors=48-2=46

E=an odd number=odd proper divisor of 2520

$\begin{array}{rcl}\mathrm{Number} \mathrm{of} \mathrm{odd} \mathrm{divisors}& =& (1+1) (2+1) (1+1)=12\\ & & \end{array}$

$\begin{array}{rcl}\mathrm{n}\left(\mathrm{E}\right)& =& 12-1=11(\mathrm{Except} 1)\end{array}$

$\begin{array}{rcl}\mathrm{P}\left(\mathrm{E}\right)& =& \frac{11}{46}\end{array}$