What is the probability that any non-leap year will have 53 Sundays?
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\textbf{Probability and Calendar.} To solve probability problems related to calendar days, consider the number of full weeks and the remaining days. The probability of a specific day occurring 53 times depends on the day of the week of the remaining days.
A non-leap year has 365 days.
We know that there are 52 weeks in a year, which accounts for \(52 \times 7 = 364\) days.
So, a non-leap year has 52 full weeks and 1 extra day.
For a non-leap year to have 53 Sundays, this extra day must be a Sunday.
The extra day can be any one of the 7 days of the week: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, or Saturday.
Each of these 7 days has an equal probability of being the extra day.
The probability that the extra day is a Sunday is 1 out of 7 possible days.
Therefore, the probability that a non-leap year will have 53 Sundays is \( \frac{1}{7} \).
