The probability that a non-leap year contains 53 Sunday is
The probability that a non-leap year contains 53 Sunday is
$\frac{1}{7}$
$\frac{1}{9}$
$\frac{2}{7}$
$\frac{1}{5}$
Solution
A non-leap year contains 365 days and there are total 52 weeks and 1 day in a non-leap year. Hence there should be 52 Sunday also.
But the 1 extra day apart from those 52 weeks may be either Monday, Tuesday, Wednesday, Thursday, Friday, Saturday or a Sunday.
Clearly, total possible events are $=7$
and Number of favourable events $=1$
Hence probability $=\frac{1}{7}$