An ordinary (non-leap) year is selected at random. What is the probability that it contains $53$ Sundays?

Step-by-step solution

1. Determine the total number of days in an ordinary year: $365$ days.
2. Convert days into weeks: $365=52\times 7+1$, which means $52$ complete weeks and $1$ extra day.
3. Observe that $52$ complete weeks will always have $52$ Sundays. The presence of a $53\text{rd}$ Sunday depends entirely on the single extra day.
4. List all possible outcomes for the extra day: $\{\text{Sunday},\text{Monday},\text{Tuesday},\text{Wednesday},\text{Thursday},\text{Friday},\text{Saturday}\}$. Total outcomes $n(S)=7$.
5. Calculate the probability of the extra day being a Sunday: $P(53\text{ Sundays})=\frac{1}{7}$.

Concept tested: Probability related to calendar years

Published 24 August 2026 · How this solution was prepared and checked

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