Ordinary (non-leap) year picked uniformly at random. What is the probability that it contains 53 Sundays?

Introduction / Context:An ordinary year has 365 days, i.e., 52 weeks plus 1 extra day. Each weekday normally appears 52 times, and one weekday appears a 53rd time. We ask the probability that Sunday is the extra occurrence.

Given Data / Assumptions:

• Ordinary year length = 52 weeks + 1 day.
• The extra day of the year is equally likely to be any of the seven weekdays for a uniformly random start day.
• We assume no bias in start-of-year weekdays.

Concept / Approach:The only way to have 53 Sundays is for the extra day to land on a Sunday.

Step-by-Step Solution:Extra day takes one of 7 weekdays with probability 1/7 each.Thus P(53 Sundays) = 1/7.

Verification / Alternative check:A leap year (366 days) has two extra days, yielding 53 of two adjacent weekdays; but here the year is ordinary, so exactly one weekday reaches 53.

Why Other Options Are Wrong:53/365 is not a valid probability model here; 2/7 would correspond to leap years, not ordinary years; other fractions do not reflect the uniform weekday assumption.

Common Pitfalls:Confusing ordinary and leap years, or assuming the calendar always starts on a fixed weekday.

Final Answer:1/7