The probability that in a year of ${22}^{nd}$ century chosen at random, there will be 53 sundays is $\frac{r}{s}$ then $r+s$ is--------(where r,s are coprime)

In the ${22}^{nd}$ century there are 25 leap years Viz 2100, 2104…..2196 and 75 non-leap years 
Consider the following events 
$E_1=$Selecting a leap year from ${22}^{nd}$ century

$E_2=$Selecting a non -leap year from ${22}^{nd}$ century

A= There are 53 Sundays in the year of ${22}^{nd}$ century

$\therefore P\left(E_1\right)=\frac{25}{100},P\left(E_2\right)=\frac{75}{100}$

$P(A/E_1)=\frac{2}{7}, P(A/E_2)=\frac{1}{7}$

Req probability $=P\left(A\right)=P\left(E_1\right)P(A/E_1)+P\left(E_2\right)P(A/E_2)=\frac{5}{28}=\frac{r}{s}\Rightarrow r+s=5+28=33$