Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Year $X$ is not a leap year. Find the probability of $X$ containing exactly 53 Sundays.

Probability

Solution:

A non-leap year has 365 days, i.e., 52 weeks and 1 day. In the first 52 weeks of that year, there will be 52 Sundays. The 1st day of the $53 rd$ week would be the last day of that year. The day of the week on this day would be the same as the day of the week on the 1st day of that year. If the first day of a non-leap year is a Sunday, that year will have 53 Sundays. Otherwise it will have 52 Sundays.
Required probability (first day of a week being a Sunday) $=\frac{1}{7}$