Find the probability that a non-leap year contains

i) 53 Sundays
ii) 52 Sundays only

Non leap year contains 365 days
365 days = 52 weeks + 1 day
That one day may be $S,M,T,W,T,F,S$

i) E be the event of 53 Sundays only Already 52 Sundays are come, one more Sunday is required

$n\left(E\right)=1\mathrm{\&}n\left(s\right)=7\Rightarrow P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{1}{7}$

ii) E be the event of 52 Sundays only .Already 52 Sundays are come extra day may be not a
Sunday

$n\left(E\right)=6$ and $n\left(S\right)=7$  $\Rightarrow P\left(E\right)=6/7$